C3 3+C3 4+C3 5+……+C3 n× C2 n+C3 n+……+Cn n=

C0 n+C1 n+C2 n+....+Cn n=2nC2 n+C3 n+.....+Cn n=2n-C0 n-C1 n Cr+Cr-1=Cr n+1 n n Cr=n!(n-r) ! r! n =C4 (n+1)-(2n-n-1)

C3 3+C3 4+C3 5+……+C3 n× C2 n+C3 n+……+Cn n=

C0 n+C1 n+C2 n+....+Cn n=2nC2 n+C3 n+.....+Cn n=2n-C0 n-C1 n Cr+Cr-1=Cr n+1 n n Cr=n!(n-r) ! r! n =C4 (n+1)-(2n-n-1)

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$({}^{3}C_{3} + {}^{4}C_{3} + {}^{5}C_{3} + \dots \dots + {}^{n}C_{3}) \times ({}^{n}C_{2} + {}^{n}C_{3} + \dots \dots + {}^{n}C_{n}) =$

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${}^{(n + 1)}C_{4} . (2^{n} - n - 1)$

${}^{(n - 1)}C_{4} . (2^{n} - n - 1)$

${}^{(n - 1)}C_{4} . (2^{n} - n + 1)$

${}^{(n - 1)}C_{4} . (2^{n} + n + 1)$

answer is A.

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Detailed Solution

$\begin{matrix} {}^{n}{C_{0}} + {}^{n}{C_{1}} + {}^{n}{C_{2}} + . . . . + {}^{n}{C_{n}} & = & 2^{n} \\ {}^{n}{C_{2}} + {}^{n}{C_{3}} + . . . . . + {}^{n}{C_{n}} & = & 2^{n} - {}^{n}{C_{0}} - {}^{n}{C_{1}} \\ {}^{n}{C_{r} + {}^{n}{C_{r - 1} = {}^{n + 1}{C_{r}}}} \\ {}^{n}{C_{r} = \frac{n!}{(n - r)! r!}} \\ = & {}^{(n + 1)}{C_{4}} - (2^{n} - n - 1) \end{matrix}$