HOW TO PART 2:
IN THE PREVIOUS POST WE WALKED THROUGH THE 6 NUMBERS – Y A I R – AND THEN BELIEF AND VALUES ETC – In that scenario we started with a situation where in you sell your house and plough the proceeds { the number A] and let it grow – with you contributing monthly an amount of $600 to the jackpot
NOW LETS SAY YOU DONT WANNA DO THAT –
IS THERE ANY OTHER SCENARIOS – PERHAPS A DIFFERENT PERMUTATION AND COMBINATION THAT CAN GET YOU TO THE SAME END POINT?
SAY YOU DONT HAVE A BIG INITIAL AMOUNT THAT YOU CAN PUT IN.. BUT CAN DO MONTHLY CONTRIBUTIONS.. ?
LUCKILY THE ANSWER IS YES..
in this scenario you don’t have to sell your house.
. all you need to do is put $20,000 now.
your monthly contribution in this scenario is $2000 – basically just deposit the Rent – Minus Taxes minus some money for buffer in a SIP – s for systematic I for Investment P for plan and viola – your FUTURE GENERATIONS ARE RICH
SEE IMAGE BELOW
al u need to remember now is the following
Y
A
I
R
NOW, YOU MAY ASK, WELL, HOW COME THIS IS ALSO SUCH A LARGE FIGURE – AND IN THIS CASE , I AM NOT EVEN HAVING TO SELL MY HOUSE?
WELL, IF U WANT TO KNOW, IN DEPTH WHY THIS IS THE CASE. , THAT BOTH THE 2 SCENARIOS – ONE IN WHICH U SELL YOUR HOUSE AND THEN INVEST – AND THE SECOND WHEREIN YOU DONT EVEN HAVE TO SELL YOUR HOUSE – BUT JUST INVEST THE RENT – LAND U IN THE SAME END POINT..
IE – BOTH GIVE U THE SAME RESULT WHICH MEANS $3.3 BILLION DOLLARS –
IF YOU REALLY WANT TO KNOW – I HAVE TO SHOW YOU THE MATHS.. IT – UNFORTUNATELY – DOES NEED AND REQUIRE SOME MATHS UNDERSTANDING AND IF U CAN VISUALIZE A NONLINEAR CURVE –
BASICALLY IN SHORT –
WE HAVE 4 VARIABLES, 4 NUMBERS,
Y A I R – THESE AREN’T FIXED,
YOU CAN CHANGE THEM – VARY THEM
– HENCE THE WORD VARIABLES –
I HAVE TO THINK ABOUT HOW TO EXPLAIN IN LAYMAN TERMS – THE MATHS –
BUT IMAGINE. F SOMEONE SAYS – ADD TWO NUMBERS – AND GET ME TO 100
HOW WOULD U DO THAT
OR IN MATHS LANGUAGE THE QUESTION IS POSED AS THE FOLLOWING
HOW MANY PERMUTATIONS AND COMBINATIONS – OF TWO NUMBERS – BOTH SAY WHOLE NUMBERS, INTEGERS – CAN BE ADDED TO MAKE THE SUM EQUAL TO 100 ?
LIKE HOW MANY
LETS SAY YOU SAY
OK
ONE IS X – AND THE SECOND IS Y – AND THEN
X PLUS Y HAS TO BE EQUAL TO 100
HOW WOULD U WRITE THAT IN MATHS NOT ENGLISH?
HERE
X + Y = 100
NOW, CAN X BE 10 AND Y BE 90 AND THEN THE RESULT BE 100 ?
OK THEN – CAN X BE 20 AND Y BE 80 AND THEN SUM BE 100?
HOW ABOUT 30 AND 70 ? – CAN X = 30 AND Y = 70 — MAKE X PLUS Y = 100 ?
OK – SO, SAY – IF SOMEONE ASKS, WHAT ARE ALL THE POSSIBLE COMBINATIONS OF X AND Y – WHERE X AND Y ARE BOTH NUMBERS – WHOLE NUMBERS NOT DECIMALS – OR FRACTIONS AND THEY WHEN ADDED EQUAL TO 100 ?
WELL, THERE WOULD BE MANY YES
HOW WOULD U KNOW HOW MANY ?
WELL, FOR THAT U WOULD NEED TO KNOW SOME MATHS – ..
U KNOW THERE ARE MANY – POSSIBILITIES, BUT IF SOMEONE ASKS, OK HOW MANY IS MANY THEN ?
WELL, THEN –
UNFORTUNATELY MATHS IN MANY CASES IS NOT LIKE ENGLISH, WHERE MANY OR VERY SUFFICES,
IN MATHS YOU HAVE TO MEASURE AND MEASUREMENT MEANS EXACT .
OK.. THAT IS NOT THAT HARD FOR THOSE WHO KNOW MATHS – AND U DONT NEED TO KNOW – ANYTHING I GUESS:) – LIBERAL ARTS MAJOR – DID U DO ? 🙂
ANYWAY, THAT WAS LINEAR – X PLUS Y = SOME Z
IS A LINEAR EQUATION..
WHERE THERE IS NO NON LINEARITY
BUT,
THE EQUATION FOR THE Y A IR – IS A NONLINEAR ONE – WHEN ITS NON LINEAR .. SO, ITS A BIT TRICKIER .
BUT,THE GENERAL IDEA I THINK U GOT – WHICH IS – THERE ARE MULTIPLE SCENARIOS – THROUGH WHICH – YOU CAN GET THE SAME RESULT.
AND HERE IS THE EQUATION
NOW, THIS IS NON LINEAR –
AND HERE – P – IS OUR A – AND R IS THE SAME – AND T IS Y.. AND A IS THE FINAL AMOUNT.
I THINK IN 2006 IN ISRAEL, U ONCE TRIED TO EXPLAIN ME – HOW INTEREST AND PRINCIPAL WORKS 🙂
THAT WAS A NICE ATTEMPT – BTW, 🙂
OK, LETS SAY U HAVE A KID.. AND BTW, MY MOM USED TO TEACH ME – EVEN WHEN SHE HERSELF DIDN’T EVEN PASS HIGH SCHOOL – LATER ONE YES SHE DID MA IN HISTORY – BUT THATS FROM LUCKNOW UNIVERSITY 🙂
OK.. SO, SAY – YOU HAVE A KID AND YOU HAVE TO TEACH
OK – MATHS – 101 – THIS IS AFTER – 1,2 3 , 4, 99, 100— TO 900000000000 TO INFINITY
IN INDIA – THIS LEVEL OF MATHS IS IN 10TH GRADE –
OK.. SO, QUESTION IS
HOW MANY DIFFERENT WAYS :
X + Y = S = 100
SO, FIRST STEP – WRITE IN MATHS – NOT IN ENGLISH
WHICH MEANS – LETS SAY – S = SUM OF X AND Y – BE EQUAL TO DENOTED BY A SYMBOL CALLED S
OK.. MATHS WORKS IN SHORT, SYMBOLS..
AND SAY – WHERE X AND Y= ANY INTEGER – { INTEGER MEANS WHOLE NUMBERS, NOT DECIMALS OR FRACTIONS}
STEP 2 – FRAME THE PROBLEM
IF S = X + Y – WHERE (X,Y BELONG TO THE FAMILY OF INTEGER NUMBERS }
THEN – HOW MANY UNIQUE – COMBINATIONS OF ( X,Y) EXIST
BTW IN MATHS – (X,Y) IS CALLED DUPLET – OR ORDERED PAIR OR SOMETHING LIKE THAT
SO, HOW WOULD YOU BEGIN –
WELL, USE YOUR BRAIN –
STEP 3 – WHAT IS BEING ASKED ? – THERE ARE TWO NUMBERS X AND Y – AND THEY CAN BE BETWEEN 0 – AND 100 – AND THEIR SUM HAS TO BE EQUAL TO PRECISELY 100.
NOW, WHY 0 TO 100 ? –
WELL,
BECAUSE – IF X = 0 AND Y = 100 – THEN X PLUS Y = 100
AND IF ANY OF THE X OR Y EXCEEDS 100 THEN THEIR SUM , S WILL EXCEED 100 ..
BUT THE PROBLEM STATES THAT S = 100 – NOT GREATER THAN OR LESS THAN BUT EQUAL TO 100
OK SO, X CAN BE BETWEEN – (0, AND 100)
AND Y CAN BE BETWEEN (0, AND 100 )
AND S = X+ Y HAS TO BE 100
ITS NOT AS EASY AS COMING UP WITH > EXAMPLES – IS IT ?
LIKE 10 PLUS 90 EQUALS 100
AND
20 PLUS 80 EQUALS 100
AND
30 PLUS 70 EQUALS 100
AND
50 PLUS 50 EQUALS 100
AND
49 PLUS 51 EQUALS TO 100
NOW U MAY SAY – THATS SO EASY – WE HAVE TWO EQUATIONS
EQ 1 – > X CAN BE BETWEEN 0 – AND 100 – THIS MEANS 100 WAYS TO CHOOSE X
EQ2 > Y CAN BE BETWEEN 0 – AND 100 – THIS MEANS 100 WAYS TO CHOOSE Y
HENCE S = SUM OF X AND Y COULD IT BE – TOTAL NUMBER OF WAYS TO CHOOSE X, COMMA Y EQUAL TO NUMBER OF WAYS TO CHOOSE X PLUS – NUMBER OF WAYS TO CHOOSE Y ?
WELL, AND THEN YOU SAY – OK THE ANSWER IS – 100 + 100 = 200 WAYS –
AND THAT WOULD BE WRONG – BECAUSE?
WELL BECAUSE – SAY X = 100 – AND Y = 100 –
THEN – S = X PLUS Y WILL BE 200 –
BUT S CANT EXCEED 100 .
OR – X = 90 AND Y = ANY NUMBER – THAT IS EITHER LESS THAN 10 – OR GREATER THAN 10 – WILL NOT RENDER S EQUAL TO 100
SEE
?
THIS IS CALLED EXAMPLES
MATHS IS –
[A]
CONCEPTUAL
[B]
ABSTRACT
AND HENCE EXAMPLES ARE FOR – SIMPLETONS 🙂
SO.. NOW, HERAD WORD BLAISE PASCAL?
NOW, IF YOU WANT TO KNOW – NOT THAT U HAVE TO KNOW – BUT IF U WANT TO KNOW – WHY MATHS IS HARD – I CAN EXPLAIN U WHY THAT IS –
ANYWAY, DID YOU GET AN IDEA THAT WHAT SEEMS LIKE VERY SIMPLE, ISNT THAT SIMPLE EITHER..
ANYWAY, THATS WHY – BANKERS – INVESTMENT BANKERS – EARN MORE MONEY – BECAUSE – IT REQUIRES THESE KINDS OF PERMUTATIONS AND COMBINATIONS WHICH IS ALL MEANS ALL – IF NOT – MOST OF WHAT BANKERS – DO – FOR THE MOST PART AS IT RELATED TO COMING UP WITH WHAT THEY CALL INNOVATIONS – WIE – NEW FINANCIAL PRODUCTS – I MEAN – ITS A START – BUT IF U GET THESE THINGS – REST IS JUST KNOWLEDGE NOT INTELLIGENCE AND VIOLA – YOUR KID COULD BECOME NVETSMENT BANKER OR ENGINEER OR SCIENTIST .
NO,W YES U DONT HAVE TO USE PASCAL’S TRIANGLE , U CAN USE MULTINOMIAL THEOREM OR IF UR LIKE ME – U CAN JUST DO YOUR OWN CALCULATION AND NOT USE ANY THEORY OR THEOREM
In Pascal’s triangle, each number is the sum of the two numbers directly above it.
In mathematics, Pascal’s triangle is a triangular array of the binomial coefficients. In much of the Western world, it is named after the French mathematician Blaise Pascal, although other mathematicians studied it centuries before him in India,^{[1]} Persia (Iran)^{[2]}, China, Germany, and Italy.^{[3]}
The rows of Pascal’s triangle are conventionally enumerated starting with row n = 0 at the top (the 0th row). The entries in each row are numbered from the left beginning with k = 0 and are usually staggered relative to the numbers in the adjacent rows. The triangle may be constructed in the following manner: In row 0 (the topmost row), there is a unique nonzero entry 1. Each entry of each subsequent row is constructed by adding the number above and to the left with the number above and to the right, treating blank entries as 0. For example, the initial number in the first (or any other) row is 1 (the sum of 0 and 1), whereas the numbers 1 and 3 in the third row are added to produce the number 4 in the fourth row.
Formula[edit]
A diagram that shows Pascal’s triangle with rows 0 through 7.
The entry in the nth row and kth column of Pascal’s triangle is denoted {\displaystyle {\tbinom {n}{k}}}. For example, the unique nonzero entry in the topmost row is {\displaystyle {\tbinom {0}{0}}=1}. With this notation, the construction of the previous paragraph may be written as follows:
 {\displaystyle {n \choose k}={n1 \choose k1}+{n1 \choose k}},
for any nonnegative integer n and any integer k between 0 and n, inclusive.^{[4]} This recurrence for the binomial coefficients is known as Pascal’s rule.
Pascal’s triangle has higher dimensional generalizations. The threedimensional version is called Pascal’s pyramid or Pascal’s tetrahedron, while the general versions are called Pascal’s simplices.
History[edit]
Yang Hui‘s triangle, as depicted by the Chinese using rod numerals, appears in a mathematical work by Zhu Shijie, dated 1303. The title reads “The Old Method Chart of the Seven Multiplying Squares” (Chinese: 古法七乘方圖; the fourth character 椉 in the image title is archaic).
Blaise Pascal’s version of the triangle
The pattern of numbers that forms Pascal’s triangle was known well before Pascal’s time. Pascal innovated many previously unattested uses of the triangle’s numbers, uses he described comprehensively in what is perhaps the earliest known mathematical treatise to be specially devoted to the triangle, his Traité du triangle arithmétique (1653). Centuries before, discussion of the numbers had arisen in the context of Indian studies of combinatorics and of binomial numbers and Greeks‘ study of figurate numbers.^{[5]}
From later commentary, it appears that the binomial coefficients and the additive formula for generating them, {\displaystyle {\tbinom {n}{r}}={\tbinom {n1}{r}}+{\tbinom {n1}{r1}}}, were known to Pingala in or before the 2nd century BC.^{[6]}^{[7]} While Pingala’s work only survives in fragments, the commentator Varāhamihira, around 505, gave a clear description of the additive formula,^{[7]} and a more detailed explanation of the same rule was given by Halayudha, around 975. Halayudha also explained obscure references to Meruprastaara, the Staircase of Mount Meru, giving the first surviving description of the arrangement of these numbers into a triangle.^{[7]}^{[8]} In approximately 850, the Jain mathematician Mahāvīra gave a different formula for the binomial coefficients, using multiplication, equivalent to the modern formula {\displaystyle {\tbinom {n}{r}}={\tfrac {n!}{r!(nr)!}}}.^{[7]} In 1068, four columns of the first sixteen rows were given by the mathematician Bhattotpala, who was the first recorded mathematician to equate the additive and multiplicative formulas for these numbers.^{[7]}
At around the same time, the Persian mathematician AlKaraji (953–1029) wrote a now lost book which contained the first description of Pascal’s triangle.^{[9]}^{[10]} It was later repeated by the Persian poetastronomermathematician Omar Khayyám (1048–1131); thus the triangle is also referred to as the Khayyam triangle in Iran.^{[11]} Several theorems related to the triangle were known, including the binomial theorem. Khayyam used a method of finding nth roots based on the binomial expansion, and therefore on the binomial coefficients.^{[12]}
Pascal’s triangle was known in China in the early 11th century through the work of the Chinese mathematician Jia Xian(1010–1070). In the 13th century, Yang Hui (1238–1298) presented the triangle and hence it is still called Yang Hui’s triangle (杨辉三角; 楊輝三角) in China.^{[13]}
In the west, the binomial coefficients were calculated by Gersonides in the early 14th century, using the multiplicative formula for them.^{[7]} Petrus Apianus (1495–1552) published the full triangle on the frontispiece of his book on business calculations in 1527. This is the first record of the triangle in Europe.^{[14]} Michael Stifel published a portion of the triangle (from the second to the middle column in each row) in 1544, describing it as a table of figurate numbers.^{[7]} In Italy, Pascal’s triangle is referred to as Tartaglia’s triangle, named for the Italian algebraist Niccolò Fontana Tartaglia (1500–1577), who published six rows of the triangle in 1556.^{[7]} Gerolamo Cardano, also, published the triangle as well as the additive and multiplicative rules for constructing it in 1570.^{[7]}
Pascal’s Traité du triangle arithmétique (Treatise on Arithmetical Triangle) was published posthumously in 1665. In this, Pascal collected several results then known about the triangle, and employed them to solve problems in probability theory. The triangle was later named after Pascal by Pierre Raymond de Montmort (1708) who called it “Table de M. Pascal pour les combinaisons” (French: Table of Mr. Pascal for combinations) and Abraham de Moivre (1730) who called it “Triangulum Arithmeticum PASCALIANUM” (Latin: Pascal’s Arithmetic Triangle), which became the modern Western name.^{[15]}
Binomial expansions[edit]
Visualisation of binomial expansion up to the 4th power
Pascal’s triangle determines the coefficients which arise in binomial expansions. For an example, consider the expansion
 (x + y)^{2} = x^{2} + 2xy + y^{2} = 1x^{2}y^{0} + 2x^{1}y^{1} + 1x^{0}y^{2}.
Notice the coefficients are the numbers in row two of Pascal’s triangle: 1, 2, 1. In general, when a binomiallike x + y is raised to a positive integer power we have:
 (x + y)^{n} = a_{0}x^{n} + a_{1}x^{n−1}y + a_{2}x^{n−2}y^{2} + … + a_{n−1}xy^{n−1} + a_{n}y^{n},
where the coefficients a_{i} in this expansion are precisely the numbers on row n of Pascal’s triangle. In other words,
 {\displaystyle a_{i}={n \choose i}.}
This is the binomial theorem.
Notice that the entire right diagonal of Pascal’s triangle corresponds to the coefficient of y^{n} in these binomial expansions, while the next diagonal corresponds to the coefficient of xy^{n−1} and so on.
To see how the binomial theorem relates to the simple construction of Pascal’s triangle, consider the problem of calculating the coefficients of the expansion of (x + 1)^{n+1} in terms of the corresponding coefficients of (x + 1)^{n} (setting y = 1 for simplicity). Suppose then that
 {\displaystyle (x+1)^{n}=\sum _{i=0}^{n}a_{i}x^{i}.}
Now
 {\displaystyle (x+1)^{n+1}=(x+1)(x+1)^{n}=x(x+1)^{n}+(x+1)^{n}=\sum _{i=0}^{n}a_{i}x^{i+1}+\sum _{i=0}^{n}a_{i}x^{i}.}
Six rows Pascal’s triangle as binomial coefficients
The two summations can be reorganized as follows:
 {\displaystyle {\begin{aligned}&\sum _{i=0}^{n}a_{i}x^{i+1}+\sum _{i=0}^{n}a_{i}x^{i}\\&{}=\sum _{i=1}^{n+1}a_{i1}x^{i}+\sum _{i=0}^{n}a_{i}x^{i}\\&{}=\sum _{i=1}^{n}a_{i1}x^{i}+\sum _{i=1}^{n}a_{i}x^{i}+a_{0}x^{0}+a_{n}x^{n+1}\\&{}=\sum _{i=1}^{n}(a_{i1}+a_{i})x^{i}+a_{0}x^{0}+a_{n}x^{n+1}\\&{}=\sum _{i=1}^{n}(a_{i1}+a_{i})x^{i}+x^{0}+x^{n+1}\end{aligned}}}
(because of how raising a polynomial to a power works, a_{0} = a_{n} = 1).
We now have an expression for the polynomial (x + 1)^{n+1} in terms of the coefficients of (x + 1)^{n} (these are the a_{i}s), which is what we need if we want to express a line in terms of the line above it. Recall that all the terms in a diagonal going from the upperleft to the lowerright correspond to the same power of x, and that the aterms are the coefficients of the polynomial (x + 1)^{n}, and we are determining the coefficients of (x + 1)^{n+1}. Now, for any given i not 0 or n + 1, the coefficient of the x^{i} term in the polynomial (x + 1)^{n+1} is equal to a_{i−1}(the figure above and to the left of the figure to be determined, since it is on the same diagonal) + a_{i} (the figure to the immediate right of the first figure). This is indeed the simple rule for constructing Pascal’s triangle rowbyrow.
It is not difficult to turn this argument into a proof (by mathematical induction) of the binomial theorem. Since (a + b)^{n} = b^{n}(a/b + 1)^{n}, the coefficients are identical in the expansion of the general case.
An interesting consequence of the binomial theorem is obtained by setting both variables x and y equal to one. In this case, we know that (1 + 1)^{n} = 2^{n}, and so
 {\displaystyle {n \choose 0}+{n \choose 1}+\cdots +{n \choose n1}+{n \choose n}=2^{n}.}
In other words, the sum of the entries in the nth row of Pascal’s triangle is the nth power of 2.
Combinations[edit]
A second useful application of Pascal’s triangle is in the calculation of combinations. For example, the number of combinations of n things taken k at a time (called n choose k) can be found by the equation
 {\displaystyle \mathbf {C} (n,k)=\mathbf {C} _{k}^{n}={_{n}C_{k}}={n \choose k}={\frac {n!}{k!(nk)!}}.}
But this is also the formula for a cell of Pascal’s triangle. Rather than performing the calculation, one can simply look up the appropriate entry in the triangle. Provided we have the first row and the first entry in a row numbered 0, the answer will be located at entry k in row n. For example, suppose a basketball team has 10 players and wants to know how many ways there are of selecting 8. The answer is entry 8 in row 10, which is 45; that is, 10 choose 8 is 45.
Relation to binomial distribution and convolutions[edit]
When divided by 2^{n}, the nth row of Pascal’s triangle becomes the binomial distribution in the symmetric case where p = 1/2. By the central limit theorem, this distribution approaches the normal distribution as n increases. This can also be seen by applying Stirling’s formula to the factorials involved in the formula for combinations.
This is related to the operation of discrete convolution in two ways. First, polynomial multiplication exactly corresponds to discrete convolution, so that repeatedly convolving the sequence {…, 0, 0, 1, 1, 0, 0, …} with itself corresponds to taking powers of 1 + x, and hence to generating the rows of the triangle. Second, repeatedly convolving the distribution function for a random variable with itself corresponds to calculating the distribution function for a sum of nindependent copies of that variable; this is exactly the situation to which the central limit theorem applies, and hence leads to the normal distribution in the limit.
Patterns and properties[edit]
Pascal’s triangle has many properties and contains many patterns of numbers.
Each frame represents a row in Pascal’s triangle. Each column of pixels is a number in binary with the least significant bit at the bottom. Light pixels represent ones and the dark pixels are zeroes.
 The sum of the elements of a single row is twice the sum of the row preceding it. For example, row 0 (the topmost row) has a value of 1, row 1 has a value of 2, row 2 has a value of 4, and so forth. This is because every item in a row produces two items in the next row: one left and one right. The sum of the elements of row n is equal to 2^{n}.
 Taking the product of the elements in each row, the sequence of products (sequence A001142 in the OEIS) is related to the base of the natural logarithm, e.^{[16]}^{[17]} Specifically, define the sequence s_{n} as follows:

 {\displaystyle s_{n}=\prod _{k=0}^{n}{\binom {n}{k}}=\prod _{k=0}^{n}{\frac {n!}{k!(nk)!}}~,~n\geq 0.}
 Then, the ratio of successive row products is
 {\displaystyle {\frac {s_{n+1}}{s_{n}}}={\frac {(n+1)!^{(n+2)}\prod _{k=0}^{n+1}{k!^{2}}}{n!^{(n+1)}\prod _{k=0}^{n}{k!^{2}}}}={\frac {(n+1)^{n}}{n!}}}
 and the ratio of these ratios is
 {\displaystyle {\frac {(s_{n+1})(s_{n1})}{(s_{n})^{2}}}=\left({\frac {n+1}{n}}\right)^{n},~n\geq 1.}
 The righthand side of the above equation takes the form of the limit definition of e
 {\displaystyle {\textit {e}}=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}.}
 Pi can be found in Pascal’s triangle through the Nilakantha infinite series.^{[18]}

 {\displaystyle \pi =3+\sum _{n=1}^{\infty }(1)^{n+1}{\frac {\binom {2n+1}{1}}{{\binom {2n+1}{2}}{\binom {2n+2}{2}}}}}
 The value of a row, if each entry is considered a decimal place (and numbers larger than 9 carried over accordingly), is a power of 11 ( 11^{n}, for row n). Thus, in row 2, ⟨1, 2, 1⟩ becomes 11^{2}, while ⟨1, 5, 10, 10, 5, 1⟩ in row five becomes (after carrying) 161,051, which is 11^{5}. This property is explained by setting x = 10 in the binomial expansion of (x + 1)^{n}, and adjusting values to the decimal system. But x can be chosen to allow rows to represent values in anybase.
 In base 3: 1 2 1_{3} = 4^{2} (16)
 ⟨1, 3, 3, 1⟩ → 2 1 0 1_{3} = 4^{3} (64)
 In base 9: 1 2 1_{9} = 10^{2} (100)
 1 3 3 1_{9} = 10^{3} (1000)
 ⟨1, 5, 10, 10, 5, 1⟩ → 1 6 2 1 5 1_{9} = 10^{5} (100000)
 In particular (see previous property), for x = 1 place value remains constant (1^{place}=1). Thus entries can simply be added in interpreting the value of a row.
 Some of the numbers in Pascal’s triangle correlate to numbers in Lozanić’s triangle.
 The sum of the squares of the elements of row n equals the middle element of row 2n. For example, 1^{2} + 4^{2} + 6^{2} + 4^{2} + 1^{2} = 70. In general form:

 {\displaystyle \sum _{k=0}^{n}{n \choose k}^{2}={2n \choose n}.}
 On any row n, where n is even, the middle term minus the term two spots to the left equals a Catalan number, specifically the (n/2 + 1)th Catalan number. For example: on row 4, 6 − 1 = 5, which is the 3rd Catalan number, and 4/2 + 1 = 3.
 In a row p where p is a prime number, all the terms in that row except the 1s are multiples of p. This can be proven easily, since if {\displaystyle p\in \mathbb {P} }, then p has no factors save for 1 and itself. Every entry in the triangle is an integer, so therefore by definition {\displaystyle (pk)!} and {\displaystyle k!} are factors of {\displaystyle p!\,}. However, there is no possible way p itself can show up in the denominator, so therefore p (or some multiple of it) must be left in the numerator, making the entire entry a multiple of p.
 Parity: To count odd terms in row n, convert n to binary. Let x be the number of 1s in the binary representation. Then the number of odd terms will be 2^{x}. These numbers are the values in Gould’s sequence.^{[19]}
 Every entry in row 2^{n}1, n ≥ 0, is odd.^{[20]}
 Polarity: When the elements of a row of Pascal’s triangle are added and subtracted together sequentially, every row with a middle number, meaning rows that have an odd number of integers, gives 0 as the result. As examples, row 4 is 1 4 6 4 1, so the formula would be 6 – (4+4) + (1+1) = 0; and row 6 is 1 6 15 20 15 6 1, so the formula would be 20 – (15+15) + (6+6) – (1+1) = 0. So every even row of the Pascal triangle equals 0 when you take the middle number, then subtract the integers directly next to the center, then add the next integers, then subtract, so on and so forth until you reach the end of the row.
Diagonals[edit]
Derivation of simplex numbers from a leftjustified Pascal’s triangle
The diagonals of Pascal’s triangle contain the figurate numbers of simplices:

 {\displaystyle {\begin{aligned}P_{0}(n)&=P_{d}(0)=1,\\P_{d}(n)&=P_{d}(n1)+P_{d1}(n)\\&=\sum _{i=0}^{n}P_{d1}(i)=\sum _{i=0}^{d}P_{i}(n1).\end{aligned}}}
The symmetry of the triangle implies that the n^{th} ddimensional number is equal to the d^{th} ndimensional number.
An alternative formula that does not involve recursion is as follows:

 {\displaystyle P_{d}(n)={\frac {1}{d!}}\prod _{k=0}^{d1}(n+k)={n^{(d)} \over d!}={\binom {n+d1}{d}}}
 where n^{(d)} is the rising factorial.
The geometric meaning of a function P_{d} is: P_{d}(1) = 1 for all d. Construct a d–dimensional triangle (a 3dimensional triangle is a tetrahedron) by placing additional dots below an initial dot, corresponding to P_{d}(1) = 1. Place these dots in a manner analogous to the placement of numbers in Pascal’s triangle. To find P_{d}(x), have a total of x dots composing the target shape. P_{d}(x) then equals the total number of dots in the shape. A 0dimensional triangle is a point and a 1dimensional triangle is simply a line, and therefore P_{0}(x) = 1 and P_{1}(x) = x, which is the sequence of natural numbers. The number of dots in each layer corresponds to P_{d − 1}(x).
Calculating a row or diagonal by itself[edit]
There are simple algorithms to compute all the elements in a row or diagonal without computing other elements or factorials.
To compute row {\displaystyle n} with the elements {\displaystyle {\tbinom {n}{0}}}, {\displaystyle {\tbinom {n}{1}}}, …, {\displaystyle {\tbinom {n}{n}}}, begin with {\displaystyle {\tbinom {n}{0}}=1}. For each subsequent element, the value is determined by multiplying the previous value by a fraction with slowly changing numerator and denominator:
 {\displaystyle {n \choose k}={n \choose k1}\times {\frac {n+1k}{k}}.}
For example, to calculate row 5, the fractions are {\displaystyle {\tfrac {5}{1}}}, {\displaystyle {\tfrac {4}{2}}}, {\displaystyle {\tfrac {3}{3}}}, {\displaystyle {\tfrac {2}{4}}} and {\displaystyle {\tfrac {1}{5}}}, and hence the elements are {\displaystyle {\tbinom {5}{0}}=1}, {\displaystyle {\tbinom {5}{1}}=1\times {\tfrac {5}{1}}=5}, {\displaystyle {\tbinom {5}{2}}=5\times {\tfrac {4}{2}}=10}, etc. (The remaining elements are most easily obtained by symmetry.)
To compute the diagonal containing the elements {\displaystyle {\tbinom {n}{0}}}, {\displaystyle {\tbinom {n+1}{1}}}, {\displaystyle {\tbinom {n+2}{2}}}, …, we again begin with {\displaystyle {\tbinom {n}{0}}=1} and obtain subsequent elements by multiplication by certain fractions:
 {\displaystyle {n+k \choose k}={n+k1 \choose k1}\times {\frac {n+k}{k}}.}
For example, to calculate the diagonal beginning at {\displaystyle {\tbinom {5}{0}}}, the fractions are {\displaystyle {\tfrac {6}{1}}}, {\displaystyle {\tfrac {7}{2}}}, {\displaystyle {\tfrac {8}{3}}}, …, and the elements are {\displaystyle {\tbinom {5}{0}}=1}, {\displaystyle {\tbinom {6}{1}}=1\times {\tfrac {6}{1}}=6}, {\displaystyle {\tbinom {7}{2}}=6\times {\tfrac {7}{2}}=21}, etc. By symmetry, these elements are equal to {\displaystyle {\tbinom {5}{5}}}, {\displaystyle {\tbinom {6}{5}}}, {\displaystyle {\tbinom {7}{5}}}, etc.
Overall patterns and properties[edit]
A level4 approximation to a Sierpinski triangle obtained by shading the first 32 rows of a Pascal triangle white if the binomial coefficient is even and black if it is odd.
 The pattern obtained by coloring only the odd numbers in Pascal’s triangle closely resembles the fractal called the Sierpinski triangle. This resemblance becomes more and more accurate as more rows are considered; in the limit, as the number of rows approaches infinity, the resulting pattern is the Sierpinski triangle, assuming a fixed perimeter.^{[21]} More generally, numbers could be colored differently according to whether or not they are multiples of 3, 4, etc.; this results in other similar patterns.
Pascal’s triangle overlaid on a grid gives the number of distinct paths to each square, assuming only rightward and downward movements are considered.
 In a triangular portion of a grid (as in the images below), the number of shortest grid paths from a given node to the top node of the triangle is the corresponding entry in Pascal’s triangle. On a Plinko game board shaped like a triangle, this distribution should give the probabilities of winning the various prizes.
 If the rows of Pascal’s triangle are leftjustified, the diagonal bands (colourcoded below) sum to the Fibonacci numbers.


1 
1 
1 
1 
2 
1 
1 
3 
3 
1 
1 
4 
6 
4 
1 
1 
5 
10 
10 
5 
1 
1 
6 
15 
20 
15 
6 
1 
1 
7 
21 
35 
35 
21 
7 
1 
Construction as matrix exponential[edit]
Binomial matrix as matrix exponential. All the dots represent 0.
Due to its simple construction by factorials, a very basic representation of Pascal’s triangle in terms of the matrix exponential can be given: Pascal’s triangle is the exponential of the matrix which has the sequence 1, 2, 3, 4, … on its subdiagonal and zero everywhere else.
Connections to geometry of polytopes[edit]
Pascal’s triangle can be used as a lookup table for the number of elements (such as edges and corners) within a polytope (such as a triangle, a tetrahedron, a square and a cube).
Number of elements of simplices[edit]
Let’s begin by considering the 3rd line of Pascal’s triangle, with values 1, 3, 3, 1. A 2dimensional triangle has one 2dimensional element (itself), three 1dimensional elements (lines, or edges), and three 0dimensional elements (vertices, or corners). The meaning of the final number (1) is more difficult to explain (but see below). Continuing with our example, a tetrahedron has one 3dimensional element (itself), four 2dimensional elements (faces), six 1dimensional elements (edges), and four 0dimensional elements (vertices). Adding the final 1 again, these values correspond to the 4th row of the triangle (1, 4, 6, 4, 1). Line 1 corresponds to a point, and Line 2 corresponds to a line segment (dyad). This pattern continues to arbitrarily highdimensioned hypertetrahedrons (known as simplices).
To understand why this pattern exists, one must first understand that the process of building an nsimplex from an (n − 1)simplex consists of simply adding a new vertex to the latter, positioned such that this new vertex lies outside of the space of the original simplex, and connecting it to all original vertices. As an example, consider the case of building a tetrahedron from a triangle, the latter of whose elements are enumerated by row 3 of Pascal’s triangle: 1 face, 3 edges, and 3 vertices (the meaning of the final 1 will be explained shortly). To build a tetrahedron from a triangle, we position a new vertex above the plane of the triangle and connect this vertex to all three vertices of the original triangle.
The number of a given dimensional element in the tetrahedron is now the sum of two numbers: first the number of that element found in the original triangle, plus the number of new elements, each of which is built upon elements of one fewer dimension from the original triangle. Thus, in the tetrahedron, the number of cells (polyhedral elements) is 0 (the original triangle possesses none) + 1 (built upon the single face of the original triangle) = 1; the number of faces is 1 (the original triangle itself) + 3 (the new faces, each built upon an edge of the original triangle) = 4; the number of edges is 3 (from the original triangle) + 3 (the new edges, each built upon a vertex of the original triangle) = 6; the number of new vertices is 3 (from the original triangle) + 1 (the new vertex that was added to create the tetrahedron from the triangle) = 4. This process of summing the number of elements of a given dimension to those of one fewer dimension to arrive at the number of the former found in the next higher simplex is equivalent to the process of summing two adjacent numbers in a row of Pascal’s triangle to yield the number below. Thus, the meaning of the final number (1) in a row of Pascal’s triangle becomes understood as representing the new vertex that is to be added to the simplex represented by that row to yield the next higher simplex represented by the next row. This new vertex is joined to every element in the original simplex to yield a new element of one higher dimension in the new simplex, and this is the origin of the pattern found to be identical to that seen in Pascal’s triangle. The “extra” 1 in a row can be thought of as the 1 simplex, the unique center of the simplex, which ever gives rise to a new vertex and a new dimension, yielding a new simplex with a new center.
Number of elements of hypercubes[edit]
A similar pattern is observed relating to squares, as opposed to triangles. To find the pattern, one must construct an analog to Pascal’s triangle, whose entries are the coefficients of (x + 2)^{Row Number}, instead of (x + 1)^{Row Number}. There are a couple ways to do this. The simpler is to begin with Row 0 = 1 and Row 1 = 1, 2. Proceed to construct the analog triangles according to the following rule:
 {\displaystyle {n \choose k}=2\times {n1 \choose k1}+{n1 \choose k}.}
That is, choose a pair of numbers according to the rules of Pascal’s triangle, but double the one on the left before adding. This results in:
1
1 2
1 4 4
1 6 12 8
1 8 24 32 16
1 10 40 80 80 32
1 12 60 160 240 192 64
1 14 84 280 560 672 448 128
The other way of manufacturing this triangle is to start with Pascal’s triangle and multiply each entry by 2^{k}, where k is the position in the row of the given number. For example, the 2nd value in row 4 of Pascal’s triangle is 6 (the slope of 1s corresponds to the zeroth entry in each row). To get the value that resides in the corresponding position in the analog triangle, multiply 6 by 2^{Position Number} = 6 × 2^{2} = 6 × 4 = 24. Now that the analog triangle has been constructed, the number of elements of any dimension that compose an arbitrarily dimensioned cube (called a hypercube) can be read from the table in a way analogous to Pascal’s triangle. For example, the number of 2dimensional elements in a 2dimensional cube (a square) is one, the number of 1dimensional elements (sides, or lines) is 4, and the number of 0dimensional elements (points, or vertices) is 4. This matches the 2nd row of the table (1, 4, 4). A cube has 1 cube, 6 faces, 12 edges, and 8 vertices, which corresponds to the next line of the analog triangle (1, 6, 12, 8). This pattern continues indefinitely.
To understand why this pattern exists, first recognize that the construction of an ncube from an (n − 1)cube is done by simply duplicating the original figure and displacing it some distance (for a regular ncube, the edge length) orthogonal to the space of the original figure, then connecting each vertex of the new figure to its corresponding vertex of the original. This initial duplication process is the reason why, to enumerate the dimensional elements of an ncube, one must double the first of a pair of numbers in a row of this analog of Pascal’s triangle before summing to yield the number below. The initial doubling thus yields the number of “original” elements to be found in the next higher ncube and, as before, new elements are built upon those of one fewer dimension (edges upon vertices, faces upon edges, etc.). Again, the last number of a row represents the number of new vertices to be added to generate the next higher ncube.
In this triangle, the sum of the elements of row m is equal to 3^{m}. Again, to use the elements of row 4 as an example: {\displaystyle 1+8+24+32+16=81}, which is equal to {\displaystyle 3^{4}=81}.
Counting vertices in a cube by distance[edit]
Each row of Pascal’s triangle gives the number of vertices at each distance from a fixed vertex in an ndimensional cube. For example, in three dimensions, the third row (1 3 3 1) corresponds to the usual threedimensional cube: fixing a vertex V, there is one vertex at distance 0 from V (that is, V itself), three vertices at distance 1, three vertices at distance √2 and one vertex at distance √3 (the vertex opposite V). The second row corresponds to a square, while largernumbered rows correspond to hypercubes in each dimension.
Fourier transform of sin(x)^{n+1}/x[edit]
As stated previously, the coefficients of (x + 1)^{n} are the nth row of the triangle. Now the coefficients of (x − 1)^{n} are the same, except that the sign alternates from +1 to −1 and back again. After suitable normalization, the same pattern of numbers occurs in the Fourier transform of sin(x)^{n+1}/x. More precisely: if n is even, take the real part of the transform, and if n is odd, take the imaginary part. Then the result is a step function, whose values (suitably normalized) are given by the nth row of the triangle with alternating signs.^{[22]} For example, the values of the step function that results from:
 {\displaystyle \,{\mathfrak {Re}}\left({\text{Fourier}}\left[{\frac {\sin(x)^{5}}{x}}\right]\right)}
compose the 4th row of the triangle, with alternating signs. This is a generalization of the following basic result (often used in electrical engineering):
 {\displaystyle \,{\mathfrak {Re}}\left({\text{Fourier}}\left[{\frac {\sin(x)^{1}}{x}}\right]\right)}
is the boxcar function.^{[23]} The corresponding row of the triangle is row 0, which consists of just the number 1.
If n is congruent to 2 or to 3 mod 4, then the signs start with −1. In fact, the sequence of the (normalized) first terms corresponds to the powers of i, which cycle around the intersection of the axes with the unit circle in the complex plane:


 {\displaystyle \,+i,1,i,+1,+i,\ldots \,}
Elementary cellular automaton[edit]
The pattern produced by an elementary cellular automaton using rule 60 is exactly Pascal’s triangle of binomial coefficients reduced modulo 2 (black cells correspond to odd binomial coefficients).^{[24]} Rule 102 also produces this pattern when trailing zeros are omitted. Rule 90 produces the same pattern but with an empty cell separating each entry in the rows.
Extensions[edit]
Pascal’s triangle can be extended to negative row numbers.
First write the triangle in the following form:

m = 0 
m = 1 
m = 2 
m = 3 
m = 4 
m = 5 
… 
n = 0 
1 
0 
0 
0 
0 
0 
… 
n = 1 
1 
1 
0 
0 
0 
0 
… 
n = 2 
1 
2 
1 
0 
0 
0 
… 
n = 3 
1 
3 
3 
1 
0 
0 
… 
n = 4 
1 
4 
6 
4 
1 
0 
… 
Next, extend the column of 1s upwards:

m = 0 
m = 1 
m = 2 
m = 3 
m = 4 
m = 5 
… 
n = −4 
1 





… 
n = −3 
1 





… 
n = −2 
1 





… 
n = −1 
1 





… 
n = 0 
1 
0 
0 
0 
0 
0 
… 
n = 1 
1 
1 
0 
0 
0 
0 
… 
n = 2 
1 
2 
1 
0 
0 
0 
… 
n = 3 
1 
3 
3 
1 
0 
0 
… 
n = 4 
1 
4 
6 
4 
1 
0 
… 
Now the rule:
 {\displaystyle {n \choose m}={n1 \choose m1}+{n1 \choose m}}
can be rearranged to:
 {\displaystyle {n1 \choose m}={n \choose m}{n1 \choose m1}}
which allows calculation of the other entries for negative rows:

m = 0 
m = 1 
m = 2 
m = 3 
m = 4 
m = 5 
… 
n = −4 
1 
−4 
10 
−20 
35 
−56 
… 
n = −3 
1 
−3 
6 
−10 
15 
−21 
… 
n = −2 
1 
−2 
3 
−4 
5 
−6 
… 
n = −1 
1 
−1 
1 
−1 
1 
−1 
… 
n = 0 
1 
0 
0 
0 
0 
0 
… 
n = 1 
1 
1 
0 
0 
0 
0 
… 
n = 2 
1 
2 
1 
0 
0 
0 
… 
n = 3 
1 
3 
3 
1 
0 
0 
… 
n = 4 
1 
4 
6 
4 
1 
0 
… 
This extension preserves the property that the values in the mth column viewed as a function of n are fit by an order m polynomial, namely
 {\displaystyle {n \choose m}={\frac {1}{m!}}\prod _{k=0}^{m1}(nk)={\frac {1}{m!}}\prod _{k=1}^{m}(nk+1)}.
This extension also preserves the property that the values in the nth row correspond to the coefficients of (1 + x)^{n}:
 {\displaystyle (1+x)^{n}=\sum _{k=0}^{\infty }{n \choose k}x^{k}\quad x<1}
For example:
 {\displaystyle (1+x)^{2}=12x+3x^{2}4x^{3}+\cdots \quad x<1}
When viewed as a series, the rows of negative n diverge. However, they are still Abel summable, which summation gives the standard values of 2^{n}. (In fact, the n = 1 row results in Grandi’s series which “sums” to 1/2, and the n = 2 row results in another wellknown series which has an Abel sum of 1/4.)
Another option for extending Pascal’s triangle to negative rows comes from extending the other line of 1s:

m = −4 
m = −3 
m = −2 
m = −1 
m = 0 
m = 1 
m = 2 
m = 3 
m = 4 
m = 5 
… 
n = −4 
1 
0 
0 
0 
0 
0 
0 
0 
0 
0 
… 
n = −3 

1 
0 
0 
0 
0 
0 
0 
0 
0 
… 
n = −2 


1 
0 
0 
0 
0 
0 
0 
0 
… 
n = −1 



1 
0 
0 
0 
0 
0 
0 
… 
n = 0 
0 
0 
0 
0 
1 
0 
0 
0 
0 
0 
… 
n = 1 
0 
0 
0 
0 
1 
1 
0 
0 
0 
0 
… 
n = 2 
0 
0 
0 
0 
1 
2 
1 
0 
0 
0 
… 
n = 3 
0 
0 
0 
0 
1 
3 
3 
1 
0 
0 
… 
n = 4 
0 
0 
0 
0 
1 
4 
6 
4 
1 
0 
… 
Applying the same rule as before leads to

m = −4 
m = −3 
m = −2 
m = −1 
m = 0 
m = 1 
m = 2 
m = 3 
m = 4 
m = 5 
… 
n = −4 
1 
0 
0 
0 
0 
0 
0 
0 
0 
0 
… 
n = −3 
−3 
1 
0 
0 
0 
0 
0 
0 
0 
0 
… 
n = −2 
3 
−2 
1 
0 
0 
0 
0 
0 
0 
0 
… 
n = −1 
−1 
1 
−1 
1 
0 
0 
0 
0 
0 
0 
.. 
n = 0 
0 
0 
0 
0 
1 
0 
0 
0 
0 
0 
… 
n = 1 
0 
0 
0 
0 
1 
1 
0 
0 
0 
0 
… 
n = 2 
0 
0 
0 
0 
1 
2 
1 
0 
0 
0 
… 
n = 3 
0 
0 
0 
0 
1 
3 
3 
1 
0 
0 
… 
n = 4 
0 
0 
0 
0 
1 
4 
6 
4 
1 
0 
… 
Note that this extension also has the properties that just as
 {\displaystyle \exp {\begin{pmatrix}.&.&.&.&.\\1&.&.&.&.\\.&2&.&.&.\\.&.&3&.&.\\.&.&.&4&.\end{pmatrix}}={\begin{pmatrix}1&.&.&.&.\\1&1&.&.&.\\1&2&1&.&.\\1&3&3&1&.\\1&4&6&4&1\end{pmatrix}},}
we have
 {\displaystyle \exp {\begin{pmatrix}.&.&.&.&.&.&.&.&.&.\\4&.&.&.&.&.&.&.&.&.\\.&3&.&.&.&.&.&.&.&.\\.&.&2&.&.&.&.&.&.&.\\.&.&.&1&.&.&.&.&.&.\\.&.&.&.&0&.&.&.&.&.\\.&.&.&.&.&1&.&.&.&.\\.&.&.&.&.&.&2&.&.&.\\.&.&.&.&.&.&.&3&.&.\\.&.&.&.&.&.&.&.&4&.\end{pmatrix}}={\begin{pmatrix}1&.&.&.&.&.&.&.&.&.\\4&1&.&.&.&.&.&.&.&.\\6&3&1&.&.&.&.&.&.&.\\4&3&2&1&.&.&.&.&.&.\\1&1&1&1&1&.&.&.&.&.\\.&.&.&.&.&1&.&.&.&.\\.&.&.&.&.&1&1&.&.&.\\.&.&.&.&.&1&2&1&.&.\\.&.&.&.&.&1&3&3&1&.\\.&.&.&.&.&1&4&6&4&1\end{pmatrix}}}
Also, just as summing along the lowerleft to upperright diagonals of the Pascal matrix yields the Fibonacci numbers, this second type of extension still sums to the Fibonacci numbers for negative index.
Either of these extensions can be reached if we define
 {\displaystyle {n \choose k}={\frac {n!}{(nk)!k!}}\equiv {\frac {\Gamma (n+1)}{\Gamma (nk+1)\Gamma (k+1)}}}
and take certain limits of the gamma function, {\displaystyle \Gamma (z)}.
SO, WHAT IS THE EASY ANSWER ?
OK HERE –
[A]
X AND Y BOTH CAN BE BETWEEN 0 AND 100
[B]
FOR EVERY X THAT YOU CHOOSE – BETWEEN O AND 100 – THERE IS ONLY ONE AND ONLY ONE POSISBLE VALUE OF Y – THAT U CAN – CHOOSE FROM BETWEEN 0 AND 100 –
[C]
SO, HOW MANY WAYS – OR COMBINATIONS ? – WELL 101 WAYS IS THE ANSWER
NOW, THIS WAS LINEAR HENCE SIMPLER, BUT WHAT IS ITS X SQUARED PLUS Y SQUARED AND WHAT IS THERE IS ANOTHER X SQUARED – THEN?
WELL, THEN HONEY, UNLESS YOU ARE RAMANUJAM – YOU WOULD HAVE TO USE ONE OF THOSE THEOREMS IN MATHS – AND THAT WOULD BE MULTI NOMIAL THEOREM
AND BTW FOR THESE KINDS OF CALCULATIONS WE HAVE – COMPUTERS – AND NOW, ON INTERNET WE HAVE FREE CALCULATORS WHICH LET U – TRY DIFFERENT PERMUTATIONS AND COMBINATIONS..
See also[edit]