# Question 64fb5

Sep 29, 2015

$n = 10$

#### Explanation:

${\text{^nC_4 = 21 * }}^{\frac{n}{2}} {C}_{3}$

So we have

 (n!)/(4! * (n-4)!) = 21 * ( (n/2)!)/(3! * [(n/2)-3]!

we have

(cancel((n-4)!) * n * (n-1) * (n-2) * (n-3))/(4! * cancel([n-4)!]) = 21 * (( cancel((n/2)-4)!) * (n/2) * [(n/2)-1] * [(n/2)-2])/(3! * [cancel((n/2)-4)!])#

$\frac{n \cdot \left(n - 1\right) \cdot \left(n - 2\right) \cdot \left(n - 3\right)}{24} = 21 \cdot \frac{\frac{n}{2} \cdot \left[\left(\frac{n}{2}\right) - 1\right] \cdot \left[\left(\frac{n}{2}\right) - 2\right]}{6}$

$\frac{n}{2}$ goes in $n$ twice and $\left(\left(\frac{n}{2}\right) - 1\right)$ goes in $\left(n - 2\right)$ twice

so we get

$\frac{\cancel{n} \cdot \left(n - 1\right) \cdot \cancel{\left(n - 2\right)} \cdot \left(n - 3\right)}{24} = 21 \cdot \frac{\cancel{n} \cdot \cancel{n - 2} \cdot \left[\left(\frac{n}{2}\right) - 2\right]}{2 \cdot 2 \cdot 6}$

$\frac{\left(n - 1\right) \cdot \left(n - 3\right)}{24} = 21 \cdot \frac{\left(\frac{n}{2}\right) - 2}{24}$

This is equivalent to

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

Next, we get

${n}^{2} - 4 \cdot n + 3 = 21 \cdot \frac{n}{2} - 42$

Bringing everything on one side we get

${n}^{2} - 14.5 \cdot n + 45 = 0$

Multiplying the whole thing by $2$ we get

$2 \cdot {n}^{2} - 29 \cdot n + 90 = 0$

$2 \cdot {n}^{2} - 20 \cdot n - 9 \cdot n + 90 = 0$

$2 \cdot n \cdot \left(n - 10\right) - 9 \cdot \left(n - 10\right) = 0$

$\left(n - 10\right) \cdot \left(2 \cdot n - 9\right) = 0$

$n = 10 \text{ }$ or $\text{ } n = 4.5$

$n$ is a natural number, so $n = 10$ is the valid answer.