# What is the value of n?

## If 2nC4: nC3 = 21:1, then find the value of n. please note that this is a permutation combination question.

Apr 23, 2018

$n = 5$

#### Explanation:

2nC4=((2n)!)/(4!(2n-4)!)

nC3=(n!)/(3!(n-3)!)

(2nC4)/(nC3)=(((2n)!)/(4!(2n-4)!))/((n!)/(3!(n-3)!))

=((2n)!3!(n-3)!)/(4!n!(2n-4)!)

Notice a few things.

(3!)/(4!)=1/4,

((2n)!)/((2n-4)!)=2n(2n-1)(2n-2)(2n-3),

and

((n-3)!)/(n!)=1/(n(n-1)(n-2)).

Putting this all together we have

((2n)!3!(n-3)!)/(4!n!(2n-4)!)=(2n(2n-1)(2n-2)(2n-3))/(4n(n-1)(n-2))

$= \frac{4 n \left(2 n - 1\right) \left(n - 1\right) \left(2 n - 3\right)}{4 n \left(n - 1\right) \left(n - 2\right)}$

$= \frac{\left(2 n - 1\right) \left(2 n - 3\right)}{n - 2}$.

We want

$\frac{\left(2 n - 1\right) \left(2 n - 3\right)}{n - 2} = 21$

$\left(2 n - 1\right) \left(2 n - 3\right) = 21 \left(n - 2\right)$

$4 {n}^{2} - 8 n + 3 = 21 n - 42$

$4 {n}^{2} - 29 n + 45 = 0$

$n = \frac{29 \pm \sqrt{{29}^{2} - 4 \left(4\right) \left(45\right)}}{2 \left(4\right)} = \frac{29 \pm 11}{8}$
So $n = 5$ or $n = \frac{9}{4}$.
Since $n$ must be an integer, $n$ must be 5.