(C) Given equation: $^{n + 1}{C_3} = 2{\,^n}{C_2}$Using the formula $^{n}{C_r} = \frac{n!}{r!(n-r)!}$,we expand both sides:$\frac{(n + 1)!}{3!(n + 1 -

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(C) Given equation: $^{n + 1}{C_3} = 2{\,^n}{C_2}$Using the formula $^{n}{C_r} = \frac{n!}{r!(n-r)!}$,we expand both sides:$\frac{(n + 1)!}{3!(n + 1 - 3)!} = 2 \times \frac{n!}{2!(n - 2)!}$$\frac{(n + 1) \times n!}{6 \times (n - 2)!} = 2 \times \frac{n!}{2 \times (n - 2)!}$Cancel $n!$ and $(n - 2)!$ from both sides:$\frac{n + 1}{6} = \frac{2}{2}$$\frac{n + 1}{6} = 1$$n + 1 = 6$$n = 5$

Competitive Quantitative Aptitude Permutation and Combination Permutation and Combination

Medium

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

A
$3$
B
$4$
C
$5$
D
$6$

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

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There are $3$ girls in a class of $10$ students. The number of different ways in which they can be seated in a row such that no two of the three girls are together is

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