If $11 [^{n - 1} c_{3}] = 24 [^{n} C_{2}]$ , then the value of n is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 11
We know that ${n_{C}}_{r} = \frac{n!}{r! (n - r)!}$
Given, $11 [{n - 1}_{C}_{3}] = 24 [{n_{C}}_{2}]$
$=> 11 \times \frac{(n - 1)!}{3! (n - 1 - 3)!} = 24 \times \frac{n!}{2! (n - 2)!}$
$=> 11 \times \frac{(n - 1)!}{3! (n - 4)!} = 24 \times \frac{n!}{2! (n - 2)!}$
$=> 11 \times \frac{(n - 1)!}{3 \times 2! (n - 4)!} = 24 \times \frac{n \times (n - 1)!}{2! (n - 2) \times (n - 3) \times (n - 4)!}$
$= . > \frac{11}{3} = \frac{24 \times n}{(n - 2) \times (n - 3)}$
$=> \frac{11}{24 \times 3} = \frac{n}{(n - 2) \times (n - 3)}$
$\frac{11}{8 \times 3 \times 3} = \frac{n}{(n - 2) \times (n - 3)}$
$=> \frac{11}{9 \times 8} = \frac{n}{(n - 2) \times (n - 3)}$
On comparing corresponding terms, we see $n = 11$

Suggest Corrections

Similar questions

10 \times^{n} C_{2} = 3 \times^{(n + 1)} C_{3}

n

If $^{n + 1} C_{3} = {2.}^{n} C_{2},$ then =

^{n + 1} C_{3} = 2^{n} C_{2}

n

^{n - 1} C_{3} +^{n - 1} C_{4} >^{n} C_{3}

n

Join BYJU'S Learning Program