Thanks in Advance,
Sincerely
Ajay
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Thanks in Advance,
Sincerely
Ajay
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[youtube http://youtube.com/w/?v=0S_UJOp9Iok]
[youtube http://youtube.com/w/?v=6qWEzgut_I4]
[youtube http://youtube.com/w/?v=K1IFCCHFANo]
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From: Streak <notifications>
Date: Fri, Apr 5, 2019 at 7:09 PM
Subject: Someone just viewed: FROM OLGA LEDNICHENKO THE RUSSHIAN IN ISARELI LOW PROTEIN BAR RAFELI CATEGORY POSTER GIRLS WHERE IS THE GARVE DIAGRSM FOR TRIP TO YONI NETANYAHU’S GRAVE STONE
To: <ajayinsead03>
Details
People on thread: YONI NETANYAHU OLGA BLOG POST BY EMAIL
Device: Android Tablet
Location: ljubljana, ?
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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.
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 I began advocating and organizing to try and change hearts and minds — and our country’s outdated and broken gun laws.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 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 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. 
If you no longer wish to receive emails from the DNC, submit this form to unsubscribe.
If you’d only like to receive our most important messages, sign up to receive less email or opt in to our SMS program by signing up to receive text messages. This email was sent to lednichenkoolga. If this isn’t your preferred email address, update your contact information. If you’re ready to elect Democrats in all 50 states, make a contribution today. This organization is powered by you, and we’d love to hear your ideas. Just reply to this email to send any comments, criticisms, or feedback. Our community management team reads every message from supporters like you. Thanks for supporting our party! Contributions or gifts to the Democratic National Committee are not tax deductible. Paid for by the Democratic National Committee, www.Democrats.org, and not authorized by any candidate or candidate’s committee. 
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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
From Wikipedia, the free encyclopedia
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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.
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}}\,,}
where
{\displaystyle {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.
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.}
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}} 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 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.}
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 }}
where
{\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}}}
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_{m1}+K=n}{n \choose k_{1},k_{2},\ldots ,k_{m1},K}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m1}^{k_{m1}}(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_{m1}+K=n}{n \choose k_{1},k_{2},\ldots ,k_{m1},K}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m1}^{k_{m1}}\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_{m1}+k_{m}+k_{m+1}=n}{n \choose k_{1},k_{2},\ldots ,k_{m1},k_{m},k_{m+1}}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m1}^{k_{m1}}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_{m1},K}{K \choose k_{m},k_{m+1}}={n \choose k_{1},k_{2},\ldots ,k_{m1},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_{m1}!K!}}{\frac {K!}{k_{m}!k_{m+1}!}}={\frac {n!}{k_{1}!k_{2}!\cdots k_{m+1}!}}.}
The numbers
{\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}}}
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}}
gives immediately that
{\displaystyle \sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}=m^{n}.}
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+m1 \choose m1}.}
The count can be performed easily using the method of stars and bars.
The largest power of a prime {\displaystyle p} that divides a multinomial coefficient may be computed using a generalization of Kummer’s theorem.
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]
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
Multiplying the number of choices at each step results in:
{\displaystyle {N \choose n_{1}}{Nn_{1} \choose n_{2}}{Nn_{1}n_{2} \choose n_{3}}\cdots ={\frac {N!}{(Nn_{1})!n_{1}!}}\cdot {\frac {(Nn_{1})!}{(Nn_{1}n_{2})!n_{2}!}}\cdot {\frac {(Nn_{1}n_{2})!}{(Nn_{1}n_{2}n_{3})!n_{3}!}}\cdots .}
Upon cancellation, we arrive at the formula given in the introduction.
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.}
(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.)
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.
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From: Dream Singles <newsletter>
Date: Fri, Apr 5, 2019 at 3:03 AM
Subject: You missed chat with Olga
To: John <ajayinsead03>
Message Olga to see if you can reschedule your chat date. Don’t miss the opportunity to talk to this beauty.

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[youtube http://youtube.com/w/?v=uM3f1iggBW8]
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