Question

Find n if
$\rm ^{n}C_{n-3}=84$

Answer 1

➡️Answer: n=9

➡️Solution:

We know that

$\rm ^{n}C_{r} = \frac{n!}{r!(n - r)!} \\ \\ here \\ \\ \rm ^{n}C_{n-3} = 84 \\ \\ \frac{n!}{(n - 3)!(n - n + 3)!} = 84 \\ \\ \frac{n!}{(n - 3)!3!} = 84 \\ \\ \frac{n(n - 1)(n - 2)}{3!} = 84 \\ \\ n(n - 1)(n - 2) = 84 \times 6 \\ \\ n( {n}^{2} - 3n + 2) = 504 \\ \\ {n}^{3} - 3 {n}^{2} + 2n - 504 = 0$
This equation's one root is calculated by hit and trial method,

On putting n=9

$= ( {9)}^{3} - 3( {9)}^{2} + 2 \times 9 - 504 \\ \\ = 729 - 243 + 18 - 504 \\ \\ = 747 - 747 \\ = 0$
by inspecting we can check that remaining roots are complex conjugate.Thus we can ignore them.

So,in the given expression value of n = 9