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Question: $^nC_0$ $^{2n}C_n$ - $^nC_1$ $^{2n-1}C_n$ + $^nC_2$ $^{2n-2}C_n$ - $^nC_3$ $^{2n-3}C_n$ + ... + ... ...

nC0^nC_0 2nCn^{2n}C_n - nC1^nC_1 2n−1Cn^{2n-1}C_n + nC2^nC_2 2n−2Cn^{2n-2}C_n - nC3^nC_3 2n−3Cn^{2n-3}C_n + ... + ... + (−1)n(-1)^n nCn^nC_n nCn^nC_n =

A

0

B

1

C

n

D

2n

Answer

1

Explanation

Solution

The given expression is a sum of products of binomial coefficients: S=∑k=0n(−1)k(nk)(2n−kn)S = \sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{2n-k}{n}

To evaluate this sum, we can use the generating function approach.
Recall that the binomial coefficient (Nr)\binom{N}{r} is the coefficient of xrx^r in the expansion of (1+x)N(1+x)^N. We denote this as [xr](1+x)N[x^r](1+x)^N.

Using this property, we can write (2n−kn)\binom{2n-k}{n} as [xn](1+x)2n−k[x^n](1+x)^{2n-k}.
Substituting this into the sum: S=∑k=0n(−1)k(nk)[xn](1+x)2n−kS = \sum_{k=0}^{n} (-1)^k \binom{n}{k} [x^n](1+x)^{2n-k} Since the coefficient operator [xn][x^n] acts linearly, we can move it outside the summation: S=[xn]∑k=0n(−1)k(nk)(1+x)2n−kS = [x^n] \sum_{k=0}^{n} (-1)^k \binom{n}{k} (1+x)^{2n-k} Now, we can factor out (1+x)2n(1+x)^{2n} from the sum: S=[xn](1+x)2n∑k=0n(−1)k(nk)(1+x)−kS = [x^n] (1+x)^{2n} \sum_{k=0}^{n} (-1)^k \binom{n}{k} (1+x)^{-k} S=[xn](1+x)2n∑k=0n(nk)(−11+x)kS = [x^n] (1+x)^{2n} \sum_{k=0}^{n} \binom{n}{k} \left(-\frac{1}{1+x}\right)^k The summation part is a binomial expansion of the form ∑k=0n(nk)akbn−k=(b+a)n\sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} = (b+a)^n.
Here, a=−11+xa = -\frac{1}{1+x} and b=1b=1. So, the sum is (1−11+x)n\left(1 - \frac{1}{1+x}\right)^n. S=[xn](1+x)2n(1+x−11+x)nS = [x^n] (1+x)^{2n} \left(\frac{1+x-1}{1+x}\right)^n S=[xn](1+x)2n(x1+x)nS = [x^n] (1+x)^{2n} \left(\frac{x}{1+x}\right)^n S=[xn](1+x)2nxn(1+x)nS = [x^n] (1+x)^{2n} \frac{x^n}{(1+x)^n} S=[xn]xn(1+x)2n−nS = [x^n] x^n (1+x)^{2n-n} S=[xn]xn(1+x)nS = [x^n] x^n (1+x)^n To find the coefficient of xnx^n in xn(1+x)nx^n (1+x)^n, we can expand (1+x)n(1+x)^n: xn(1+x)n=xn∑j=0n(nj)xj=∑j=0n(nj)xn+jx^n (1+x)^n = x^n \sum_{j=0}^{n} \binom{n}{j} x^j = \sum_{j=0}^{n} \binom{n}{j} x^{n+j} We are looking for the term where the power of xx is nn. This occurs when n+j=nn+j = n, which implies j=0j=0.
The coefficient corresponding to j=0j=0 is (n0)\binom{n}{0}.
Since (n0)=1\binom{n}{0} = 1, the sum SS is equal to 1.

This result can also be confirmed by a known combinatorial identity:
∑k=0n(−1)k(nk)(m−kr)=(m−nr−n)\sum_{k=0}^{n} (-1)^k \binom{n}{k} \binom{m-k}{r} = \binom{m-n}{r-n}.
In our problem, m=2nm=2n and r=nr=n.
Substituting these values:
S=(2n−nn−n)=(n0)=1S = \binom{2n-n}{n-n} = \binom{n}{0} = 1.