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I am studying for an upcoming test and I was having trouble with this practice problem:

Determine the coefficient of $x^{111}y^{444}$ in the expansion of $(17x + 71y)^{555}$.

I am thinking of using the binomial theorem but I do not think I am on the right track. Any help would be much appreciated. Thanks!

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    $\begingroup$ yes, you should use the binomial theorem $\endgroup$ – Leox Sep 29 '13 at 18:48
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The $r$th term of $\displaystyle(a+b)^n$ is $$\binom nr a^{n-r}b^r$$

So, the $r$th term of $(17x+71y)^{555}$ is $$\binom{555}r (17x)^{555-r}(71y)^r=\binom{555}r (17)^{(555-r)}(71)^r x^{(555-r)}y^r$$

Can you take it from here?

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  • $\begingroup$ The initial equation makes sense but I'm still having trouble wrapping my mind around what r is. Can you elaborate please? $\endgroup$ – pyramid Sep 29 '13 at 19:59
  • $\begingroup$ @pyramid, as the binomial expansion with positive integer$(n)$ contains $n+1$ terms, integer $r$ can assume values between $[0,n]$ $\endgroup$ – lab bhattacharjee Sep 30 '13 at 5:59
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By the way, I don't think you are expected to find the numerical answer, which is $$ 60064047680957680768646960586906579644863861729947868363421725532536245655816580\\959186095590614743032901549098483041058692039487532976027431680927788189258884\\667921934897208160966434297204970367725257673678334640020467536376922367772375\\238614797303560225331118522523871027401717663456479612623850394930029576518799\\173213346032944950454451556549947575246030772311815866610463738530270970312075\\742463132636608194497892880666643437549504349031121810430324920022844269880174\\152100324459472673932649711930826632466357408146364537972489521949348293986570\\193110219185231021279261370670588231426471595469889861269779151924494325125060\\959053916780578580280731646135881843879230223490128953500103456424990049870766\\674356454814657650873779679145586907602736046617845378999159321662525694629969\\220854145761549112400110003859948555752184106292665646129788470041179713084014\\621603890129540681451973397700120434864808660249953975352052984480553471180447\\622187492766682155506802536865986592663604223327228955439785871964767368379537\\18033688958209159665744154952422387117235549344467984825488000$$

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$$(a+b)^n=\binom{n}{0}a^0b^n+\binom{n}{1}a^1b^{n-1}+\binom{n}{2}a^2b^{n-2}+\cdots+\binom{n}{n}a^nb^0=\sum_{k=0}^{k=n}\binom{n}{k}a^kb^{n-k}$$

There is a clear combinatorial meaning. We pick $a$ or $b$ from $a+b$ and repeat $n$ times. The coefficient of $a^kb^{n-k}$ means that we pick $k$ $a$'s from $n$ $a$'s,i.e. $\binom{n}{k}$. Perhaps it helps to understand.

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