Find the coefficient of x^5 in (1+x)^11.

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$According to the binomial theorem$

$(1+x)^n= \displaystyle\sum_{k=0}^{n} \binom{n}{k}x^k$

$\text{coefficient of }x^5 \ \binom{11}{5}=\frac{11!}{5!6!}=462$

Answer: 462

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Question ID: mtid-5-stid-8-sqid-3622-qpid-2321