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Can the following sum be reduced to an analytic expression?

$$\sum_{m=0}^N\frac{N!}{m!(N-m)!}$$

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closed as off-topic by callculus, NCh, Andrew, max_zorn, saz Jan 30 at 7:52

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    $\begingroup$ Maybe $2^N$ - I think. Apply the binomial theorem. Give a reply if something is still unclear. $\endgroup$ – callculus Jan 27 at 16:20
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    $\begingroup$ This sums is equal to the number of subsets of a set with $N$ elements, therefore it is $2^N.$ $\endgroup$ – user376343 Jan 27 at 16:24
  • $\begingroup$ For future questions, don't jump to SE before working yourself. Here you could try evaluating this for $N=1,2,3$, maybe $4$, Then you could guess the pattern and try to prove it. $\endgroup$ – Ethan Bolker Jan 27 at 16:37
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Note: $$(1+x)^n= \ ^nC_0.1^n.x^0+^nC_1.1^{n-1}.x^1+^nC_2.1^{n-2}.x^2+^nC_3.1^{n-3}.x^3+\ldots +^nC_n.1^0.x^n$$

Put x = 1, to recover $$\sum_{m=0}^{n} \ {^n}C_m = 2^n$$

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