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I’m Jewish. I’m Middle Eastern. I’m an immigrant. My life and my community have been shaped by gun violence. And now I’m organizing to help retake the White House and better our country.

SHALOM AMIT

WELL, GUNS WHEN ACCESSES IS SO EASY IN A VIOLENT AGGRESSIVE CULTURE – THEN WE HAVE –

[A] RITUALS – AND CEMEMORNICAL SPORTS – THIS MEANS SCHOOOL SHOOTTINGS

AND

[B] HONORARRY – SPORTS – THIS – U PROBABLY KNOW LIKE ME DO BEING JEWISH –

BTW, JEWS AND SOMEHOW HINDUS ARENT FOUD HAVING A BURNING DESIRE TO DO THIS SPORTS ?

SOMEHOW, THATS – THE CASE – ISNT IT AMIT ?

BTW AMIT IS ALSO A HINDU NAME – DO U KNOW ?

REGARDS AND SHALOM NAMASTE

OLYA – SHULMAN –

From: Amit Dadon <democraticparty>
Date: Fri, Apr 5, 2019 at 9:30 PM
Subject: I’m Jewish. I’m Middle Eastern. I’m an immigrant. My life and my community have been shaped by gun violence. And now I’m organizing to help retake the White House and better our country.
To: <lednichenkoolga>

Since I got involved in the fight to end gun violence over a year ago, I’ve come to realize just how much of a difference we as young people can make when we organize together.

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I never thought I’d be an organizer for gun violence prevention, but that was before gun violence claimed the lives of 17 students at my alma mater in Parkland, Florida.

The massacre that day devastated me. It devastated my community, and forever changed the lives of the thousands affected. Horrified, I felt the need to take action. I began advocating and organizing to try and change hearts and minds — and our country’s outdated and broken gun laws. Now, I’m proud to be taking the next step by joining Organizing Corps 2020 to help elect a Democrat to the White House.

As a minority, immigrant, and young person, I spent too long thinking my voice couldn’t make a difference. Over the past 14 months, I’ve realized that it can, and that’s what makes me so excited about Organizing Corps 2020.

The DNC is helping train people like me — people from groups too often underrepresented in politics — to be leaders in their community in ways that will continue to have an impact long after the 2020 election.

But here’s the catch: Organizing Corps 2020 isn’t fully funded yet, and right now we need dedicated Democrats like you to help support it.

Can you make a $3 donation today to support crucial DNC programs like Organizing Corps 2020? Your early support will make a big difference in ensuring that young organizers like me receive the training we need to hit the ground running in 2020.

DONATE: $3
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Or donate another amount.

Since I got involved in the fight to end gun violence, I’ve realized how we can truly make a difference if we come together to fight and organize for a common purpose. In the past 14 months, two gun violence prevention bills have passed in the House, state legislatures nationwide have passed and enacted common sense gun laws, and public opinion is shifting for the better — but we’re just getting started.

The progress we’ve made on gun violence prevention is proof that a better world is possible, and we must empower people with the tools and skills they need to create change in their communities, states, and this country as a whole. Organizing Corps can play a vital role in making that happen — starting with the 2020 elections — but we need your help.

I hope you’ll donate $3 today to support important DNC programs like Organizing Corps 2020.

Thanks for believing in us.

Amit

Amit Dadon
University of Maryland, 2020
Organizing Corps 2020 Fellow

P.S. The deadline to apply for Organizing Corps 2020 is Monday at midnight! If you have any young, passionate people in your life who are ready to help elect our next Democratic president (and get paid while doing it), send them this link before it’s too late.

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TO BIBI NETANYAHU FROM OLGA SHULMAN LEDNICHENKO AND ME -> THE ISARELI LIST ARITHMETIC AND BLOCK CHALENGE – WHICH IS AN EXCERCISE FOR YOU MR PRIME MINISTER – WHO IS ABOUT TO BREAK THE BEN GURION RECORD -> SO ? BIBI -> AFTER VICTORY AND RESULTS IMEEXPLAIN HOW YONI NETANYAHU DOES ISARELI PREDICTIONS VIA ARITHEMTICS – SO, OLGA NETANAYHU MATCH AND CHALLENGE

HERE IS THE ANSWER SOMEWHERE

NOW, BIBI, ME RUSSIAN – AND U BORN IN TEL AVIV

SO, HERE IS THE ANSWER

4
NOW, BIBI, U WOULD HAVE SOME BALL AND PARTY – AND HERE IS MY PARTY

HINT – U WILL HAVE TO – APPLY – THIS – MULTI NOMILAL THEOREM ON YOUR TAL LIST AND THE LESFFT BLOCK AND THE CENTRIST AND OTHER BLOCKS

Multinomial theorem

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In mathematics, the multinomial theorem describes how to expand a power of a sum in terms of powers of the terms in that sum. It is the generalization of the binomial theorem from binomials to multinomials.

Contents

Theorem[edit]

For any positive integer m and any nonnegative integer n, the multinomial formula tells us how a sum with m terms expands when raised to an arbitrary power n:

{\displaystyle (x_{1}+x_{2}+\cdots +x_{m})^{n}=\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}\prod _{t=1}^{m}x_{t}^{k_{t}}\,,}{\displaystyle (x_{1}+x_{2}+\cdots +x_{m})^{n}=\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}\prod _{t=1}^{m}x_{t}^{k_{t}}\,,}

where

{\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}}{n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}

is a multinomial coefficient. The sum is taken over all combinations of nonnegative integer indices k1 through km such that the sum of all ki is n. That is, for each term in the expansion, the exponents of the xi must add up to n. Also, as with the binomial theorem, quantities of the form x0 that appear are taken to equal 1 (even when x equals zero).

In the case m = 2, this statement reduces to that of the binomial theorem.

Example[edit]

The third power of the trinomial a + b + c is given by

{\displaystyle (a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3a^{2}b+3a^{2}c+3b^{2}a+3b^{2}c+3c^{2}a+3c^{2}b+6abc.}(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3a^{2}b+3a^{2}c+3b^{2}a+3b^{2}c+3c^{2}a+3c^{2}b+6abc.

This can be computed by hand using the distributive property of multiplication over addition, but it can also be done (perhaps more easily) with the multinomial theorem, which gives us a simple formula for any coefficient we might want. It is possible to "read off" the multinomial coefficients from the terms by using the multinomial coefficient formula. For example:

{\displaystyle a^{2}b^{0}c^{1}}a^{2}b^{0}c^{1} has the coefficient {\displaystyle {3 \choose 2,0,1}={\frac {3!}{2!\cdot 0!\cdot 1!}}={\frac {6}{2\cdot 1\cdot 1}}=3.}{\displaystyle {3 \choose 2,0,1}={\frac {3!}{2!\cdot 0!\cdot 1!}}={\frac {6}{2\cdot 1\cdot 1}}=3.}{\displaystyle a^{1}b^{1}c^{1}}a^{1}b^{1}c^{1} has the coefficient {\displaystyle {3 \choose 1,1,1}={\frac {3!}{1!\cdot 1!\cdot 1!}}={\frac {6}{1\cdot 1\cdot 1}}=6.}{\displaystyle {3 \choose 1,1,1}={\frac {3!}{1!\cdot 1!\cdot 1!}}={\frac {6}{1\cdot 1\cdot 1}}=6.}

Alternate expression[edit]

The statement of the theorem can be written concisely using multiindices:

{\displaystyle (x_{1}+\cdots +x_{m})^{n}=\sum _{|\alpha |=n}{n \choose \alpha }x^{\alpha }}(x_{1}+\cdots +x_{m})^{n}=\sum _{{|\alpha |=n}}{n \choose \alpha }x^{\alpha }

where

{\displaystyle \alpha =(\alpha _{1},\alpha _{2},\dots ,\alpha _{m})}{\displaystyle \alpha =(\alpha _{1},\alpha _{2},\dots ,\alpha _{m})}

and

{\displaystyle x^{\alpha }=x_{1}^{\alpha _{1}}x_{2}^{\alpha _{2}}\cdots x_{m}^{\alpha _{m}}}{\displaystyle x^{\alpha }=x_{1}^{\alpha _{1}}x_{2}^{\alpha _{2}}\cdots x_{m}^{\alpha _{m}}}

Proof[edit]

This proof of the multinomial theorem uses the binomial theorem and induction on m.

First, for m = 1, both sides equal x1n since there is only one term k1 = n in the sum. For the induction step, suppose the multinomial theorem holds for m. Then

{\displaystyle {\begin{aligned}&(x_{1}+x_{2}+\cdots +x_{m}+x_{m+1})^{n}=(x_{1}+x_{2}+\cdots +(x_{m}+x_{m+1}))^{n}\\[6pt]={}&\sum _{k_{1}+k_{2}+\cdots +k_{m-1}+K=n}{n \choose k_{1},k_{2},\ldots ,k_{m-1},K}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m-1}^{k_{m-1}}(x_{m}+x_{m+1})^{K}\end{aligned}}}{\displaystyle {\begin{aligned}&(x_{1}+x_{2}+\cdots +x_{m}+x_{m+1})^{n}=(x_{1}+x_{2}+\cdots +(x_{m}+x_{m+1}))^{n}\\[6pt]={}&\sum _{k_{1}+k_{2}+\cdots +k_{m-1}+K=n}{n \choose k_{1},k_{2},\ldots ,k_{m-1},K}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m-1}^{k_{m-1}}(x_{m}+x_{m+1})^{K}\end{aligned}}}

by the induction hypothesis. Applying the binomial theorem to the last factor,

{\displaystyle =\sum _{k_{1}+k_{2}+\cdots +k_{m-1}+K=n}{n \choose k_{1},k_{2},\ldots ,k_{m-1},K}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m-1}^{k_{m-1}}\sum _{k_{m}+k_{m+1}=K}{K \choose k_{m},k_{m+1}}x_{m}^{k_{m}}x_{m+1}^{k_{m+1}}}=\sum _{{k_{1}+k_{2}+\cdots +k_{{m-1}}+K=n}}{n \choose k_{1},k_{2},\ldots ,k_{{m-1}},K}x_{1}^{{k_{1}}}x_{2}^{{k_{2}}}\cdots x_{{m-1}}^{{k_{{m-1}}}}\sum _{{k_{m}+k_{{m+1}}=K}}{K \choose k_{m},k_{{m+1}}}x_{m}^{{k_{m}}}x_{{m+1}}^{{k_{{m+1}}}}{\displaystyle =\sum _{k_{1}+k_{2}+\cdots +k_{m-1}+k_{m}+k_{m+1}=n}{n \choose k_{1},k_{2},\ldots ,k_{m-1},k_{m},k_{m+1}}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m-1}^{k_{m-1}}x_{m}^{k_{m}}x_{m+1}^{k_{m+1}}}=\sum _{{k_{1}+k_{2}+\cdots +k_{{m-1}}+k_{m}+k_{{m+1}}=n}}{n \choose k_{1},k_{2},\ldots ,k_{{m-1}},k_{m},k_{{m+1}}}x_{1}^{{k_{1}}}x_{2}^{{k_{2}}}\cdots x_{{m-1}}^{{k_{{m-1}}}}x_{m}^{{k_{m}}}x_{{m+1}}^{{k_{{m+1}}}}

which completes the induction. The last step follows because

{\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{m-1},K}{K \choose k_{m},k_{m+1}}={n \choose k_{1},k_{2},\ldots ,k_{m-1},k_{m},k_{m+1}},}{n \choose k_{1},k_{2},\ldots ,k_{{m-1}},K}{K \choose k_{m},k_{{m+1}}}={n \choose k_{1},k_{2},\ldots ,k_{{m-1}},k_{m},k_{{m+1}}},

as can easily be seen by writing the three coefficients using factorials as follows:

{\displaystyle {\frac {n!}{k_{1}!k_{2}!\cdots k_{m-1}!K!}}{\frac {K!}{k_{m}!k_{m+1}!}}={\frac {n!}{k_{1}!k_{2}!\cdots k_{m+1}!}}.}{\frac {n!}{k_{1}!k_{2}!\cdots k_{{m-1}}!K!}}{\frac {K!}{k_{m}!k_{{m+1}}!}}={\frac {n!}{k_{1}!k_{2}!\cdots k_{{m+1}}!}}.

Multinomial coefficients[edit]

The numbers

{\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{m}}}{\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{m}}}

appearing in the theorem are the multinomial coefficients. They can be expressed in numerous ways, including as a product of binomial coefficients or of factorials:

{\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}={k_{1} \choose k_{1}}{k_{1}+k_{2} \choose k_{2}}\cdots {k_{1}+k_{2}+\cdots +k_{m} \choose k_{m}}}{\displaystyle {n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}={k_{1} \choose k_{1}}{k_{1}+k_{2} \choose k_{2}}\cdots {k_{1}+k_{2}+\cdots +k_{m} \choose k_{m}}}

Sum of all multinomial coefficients[edit]

The substitution of xi = 1 for all i into the multinomial theorem

{\displaystyle \sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m}^{k_{m}}=(x_{1}+x_{2}+\cdots +x_{m})^{n}}{\displaystyle \sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m}^{k_{m}}=(x_{1}+x_{2}+\cdots +x_{m})^{n}}

gives immediately that

{\displaystyle \sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}=m^{n}.}{\displaystyle \sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}=m^{n}.}

Number of multinomial coefficients[edit]

The number of terms in a multinomial sum, #n,m, is equal to the number of monomials of degree n on the variables x1, …, xm:

{\displaystyle \#_{n,m}={n+m-1 \choose m-1}.}{\displaystyle \#_{n,m}={n+m-1 \choose m-1}.}

The count can be performed easily using the method of stars and bars.

Valuation of multinomial coefficients[edit]

The largest power of a prime {\displaystyle p}p that divides a multinomial coefficient may be computed using a generalization of Kummer’s theorem.

Interpretations[edit]

Ways to put objects into bins[edit]

The multinomial coefficients have a direct combinatorial interpretation, as the number of ways of depositing n distinct objects into m distinct bins, with k1 objects in the first bin, k2 objects in the second bin, and so on.[1]

Number of ways to select according to a distribution[edit]

In statistical mechanics and combinatorics if one has a number distribution of labels then the multinomial coefficients naturally arise from the binomial coefficients. Given a number distribution {ni} on a set of N total items, ni represents the number of items to be given the label i. (In statistical mechanics i is the label of the energy state.)

The number of arrangements is found by

  • Choosing n1 of the total N to be labeled 1. This can be done {\displaystyle N \choose n_{1}}{N \choose n_{1}} ways.
  • From the remaining Nn1 items choose n2 to label 2. This can be done {\displaystyle N-n_{1} \choose n_{2}}{N-n_{1} \choose n_{2}} ways.
  • From the remaining Nn1 − n2 items choose n3 to label 3. Again, this can be done {\displaystyle N-n_{1}-n_{2} \choose n_{3}}{N-n_{1}-n_{2} \choose n_{3}} ways.

Multiplying the number of choices at each step results in:

{\displaystyle {N \choose n_{1}}{N-n_{1} \choose n_{2}}{N-n_{1}-n_{2} \choose n_{3}}\cdots ={\frac {N!}{(N-n_{1})!n_{1}!}}\cdot {\frac {(N-n_{1})!}{(N-n_{1}-n_{2})!n_{2}!}}\cdot {\frac {(N-n_{1}-n_{2})!}{(N-n_{1}-n_{2}-n_{3})!n_{3}!}}\cdots .}{\displaystyle {N \choose n_{1}}{N-n_{1} \choose n_{2}}{N-n_{1}-n_{2} \choose n_{3}}\cdots ={\frac {N!}{(N-n_{1})!n_{1}!}}\cdot {\frac {(N-n_{1})!}{(N-n_{1}-n_{2})!n_{2}!}}\cdot {\frac {(N-n_{1}-n_{2})!}{(N-n_{1}-n_{2}-n_{3})!n_{3}!}}\cdots .}

Upon cancellation, we arrive at the formula given in the introduction.

Number of unique permutations of words[edit]

The multinomial coefficient is also the number of distinct ways to permute a multiset of n elements, and ki are the multiplicitiesof each of the distinct elements. For example, the number of distinct permutations of the letters of the word MISSISSIPPI, which has 1 M, 4 Is, 4 Ss, and 2 Ps is

{\displaystyle {11 \choose 1,4,4,2}={\frac {11!}{1!\,4!\,4!\,2!}}=34650.}{11 \choose 1,4,4,2}={\frac {11!}{1!\,4!\,4!\,2!}}=34650.

(This is just like saying that there are 11! ways to permute the letters—the common interpretation of factorial as the number of unique permutations. However, we created duplicate permutations, because some letters are the same, and must divide to correct our answer.)

Generalized Pascal’s triangle[edit]

One can use the multinomial theorem to generalize Pascal’s triangle or Pascal’s pyramid to Pascal’s simplex. This provides a quick way to generate a lookup table for multinomial coefficients.

See also

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