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Let $P_{n}$ denote the product of all the coefficients in the expansion of expansion of ${(1 + x)}^{n} .$ If $(20)! P_{n + 1} = 21^{20} P_{n},$ then $n$ is equal to

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Correct option is B

Suppose (1+x)n+1=∑k=0n+1 Bkxk, where Bk=n+1Ckand (1+x)n=∑k=0n Ckxk, where Ck=nCk,We have Pn+1Pn=B¯0B1A0B2A1⋯Bn+1AnNow, Br+1Ar=(n+1)!(r+1)!(n−r)!⋅r!(n−r)!n!=n+1r+1 (0≤r≤n)Thus, Pn+1Pn=(n+1)nr!⇒212020!=(n+1)nr!⇒n=20

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Let Pn denote the product of all the coefficients in the expansion of expansion of (1+x)n. If (20)!Pn+1=2120Pn, then n is equal to

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Let $P_{n}$ denote the product of all the coefficients in the expansion of expansion of ${(1 + x)}^{n} .$ If $(20)! P_{n + 1} = 21^{20} P_{n},$ then $n$ is equal to