Consider: $(a + b + c + d)^{21}$
What will be the coefficient of $a^5b^6c^5d^5$?
Can you also give the general way of finding out coefficients of such terms?
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Do you have only one sum to the power of some natural number ? – Peter Sep 27 '16 at 15:24
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Since $5+6+5+5\neq 2$ the coefficient is zero. – guestDiego Sep 27 '16 at 15:24
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@guestDiego You mean $5+6+5+5>2$. – Peter Sep 27 '16 at 15:24
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@Peter - I have edited the question, had made a mistake. – crimsonKnight Sep 27 '16 at 15:24
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1Try for the coefficient of $a^5 b^6 c^5 d^5$ in $(a+b+c+d)^{21}$. – Robert Israel Sep 27 '16 at 15:25
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1@Peter The coefficient would be zero also if $n_a+n_b+n_c+n_d=1$ – guestDiego Sep 27 '16 at 15:26
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21- there you go – crimsonKnight Sep 27 '16 at 15:26
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This is just a application of the multinomial theorem. I'm sure you could find an another question about this. – S.C.B. Sep 27 '16 at 15:27
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See: https://en.wikipedia.org/wiki/Multinomial_theorem – Emilio Novati Sep 27 '16 at 15:28
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First of all, it should be clarified whether the question is about a single power. – Peter Sep 27 '16 at 15:29
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Related. – Sep 27 '16 at 15:30
1 Answers
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$a^5b^6c^5d^5$ corresponds to the coefficient $$\frac{21!}{5!6!5!5!}$$ in the multinomial $(a+b+c+d)^{21}$. The corresponding value is $\color{red}{41\space 064\space 607\space 584}$.
Piquito
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So, does $a^7b^5c^8d^4$ in $(a + b + c + d)^{21}$ has a coefficient of $\frac{21!}{7!5!8!4!}$? – crimsonKnight Sep 29 '16 at 14:38
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And I had another confusion: is it necessary that the powers of a, b, c and d must sum up to the x in $(a + b + c + d)^x$? – crimsonKnight Sep 29 '16 at 14:46
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Always, like for Newton Binomial. You should get an homogeneous polynomial, so all the terms of the sum have same degree. – Piquito Sep 30 '16 at 19:01
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Respect to the coefficients, all the terms of the sum have the form $a^{n_1}b^{n_2}c^{n_3}d^{n_4}$ where $n_1+n_2+n_3+n_4=21$ (the same happens with the binomial). – Piquito Sep 30 '16 at 19:13