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Let $n$ be a positive integer, and let $a, b$ be integers greater than or equal to 0 such that $a+b\le n$. Determine the coefficient of $x^ay^b$ in the expansion of $(1+x+y)^n$. Give a counting argument for your answer.

I am aware that the question is about binomial coefficients but I'm not too sure how to go about this problem. Any help would be appreciated.

Haxify
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2 Answers2

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You can make a counting argument; if you expand the expression

$$(1+x+y)(1+x+y)...(1+x+y)$$

First you need to choose $x$ from $a$ of the parenthesized units, which you can do in $\binom n a$ ways. Then you select $b$ of the $y$s from the remaining, in $\binom {n-a} b$, and the rest are fixed as $1$.

$$\textrm{# of ways to select} = \textrm{coefficient} = \binom n a \binom {n-a} b$$

  • Hi, thank you for the answer. This may be a really stupid question, but how does the # of ways to select become the coefficient for the term after the expansion? I'm still new to this type of questions, sorry! – Haxify Jan 14 '16 at 23:43
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    So when you expand the expression, you pick one term from each as you expand it (sort of like FOIL; first you expand $(a+b)(c+d)$ by picking $a$ and multiplying by $c,d$, then by picking $b$ and multiplying by $c,d$). –  Jan 14 '16 at 23:50
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    When you expand the expression, you are actually looking at every possible product with a member from each of the parenthesized terms, and this counts the number of products that will involve $a$ of the $x$s and $b$ of the $y$s, with the rest all $1$s (which gives an $x^ay^b$ term) –  Jan 14 '16 at 23:52
  • Oh, I get it now. Thank you so much! – Haxify Jan 14 '16 at 23:52
  • You're welcome! This is actually a pretty cool problem - you will see things like this come back as 'generating functions' in later math classes. –  Jan 15 '16 at 00:01
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Another way is select first the $a+b$ factors and, over these $a+b$ choose de $a$ factors for $x$: $\binom{n}{a+b}\binom{a+b}{a}$ (indeed, both arguments shows the identity $\binom{n}{a+b}\binom{a+b}{a}=\binom{n}{a}\binom{n-a}{b}$).

sinbadh
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