Q.152. Binomial Theorem and combinations

Question 152.
If (1 + x)^n = C0 + C1x + C2x^2 + ...... + Cn,  then  prove that
C0Cn+C1Cn-1+C2Cn-2+.....+CnC0 = (2n)! / (n!) (n!).

Answer 152.
(1+x)^n = C0+C1x+C2x^2+......+Cnx^n
In [ (1+x)^n * (1+x)^n ], the coefficient of x^n is
C0Cn+C1Cn-1+C2Cn-2+.....+CnC0
But, [ (1+x)^n * (1+x)^n ] = (1+x)^(2n)
In the binomial expansion of (1+x)^(2n), coefficient of x^(n) is (2n)Cn
Therefore,
C0Cn+C1Cn-1+C2Cn-2+.....+CnC0 = (2n)! / (n!) (n!).

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