If ^{n + 1}{C₃} = 2{,^n}{C₂}, then n =

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6.Permutation and Combination

easy

If $^{n + 1}{C_3} = 2{\,^n}{C_2},$ then $n =$

A

$3$

B

$4$

C

$5$

D

$6$

Solution

(c) $^{n + 1}{C_3} = 2\,.{\,^n}{C_2}$

==> $\frac{{(n + 1)!}}{{3!\,.\,(n – 2)!}} = 2\,.\,\frac{{n!}}{{2!.(n – 2)!}}$

==> $\frac{{n + 1}}{{3\,.\,2!}} = \frac{2}{{2!}}$

==> $n + 1 = 6\,\, \Rightarrow \,n = 5$.

Standard 11

Mathematics

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(IIT-1979)