Question

If ${}^n{C_4} = {}^n{C_6}$ , find ${}^{12}{C_n}$.

Answer 1

We can expand the given equation using the expansion of combinations. Then we can simplify them using properties of factorials. Then we will obtain a quadratic equation. On solving we can obtain the value of n, then we can substitute the value of n in ${}^{12}{C_n}$ and find its value using the expansion of combinations.

Complete step by step solution:
We are given that ${}^n{C_4} = {}^n{C_6}$ .
We know that, the expansion of the combinations ${}^n{C_r}$ is given by,
${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
On using this relation on the given equation, we get,
$\Rightarrow \dfrac{{n!}}{{4!\left( {n - 4} \right)!}} = \dfrac{{n!}}{{6!\left( {n - 6} \right)!}}$
On cancelling the common terms, we get,
$\Rightarrow \dfrac{1}{{4!\left( {n - 4} \right)!}} = \dfrac{1}{{6!\left( {n - 6} \right)!}}$
On cross multiplying, we get,
$\Rightarrow \dfrac{{6!}}{{4!}} = \dfrac{{\left( {n - 4} \right)!}}{{\left( {n - 6} \right)!}}$
We know that $n! = n \times \left( {n - 1} \right)!$ . On using this relation, we get,
$\Rightarrow \dfrac{{6 \times 5 \times 4!}}{{4!}} = \dfrac{{\left( {n - 4} \right) \times \left( {n - 5} \right) \times \left( {n - 6} \right)!}}{{\left( {n - 6} \right)!}}$
On cancelling the common terms, we get,
$\Rightarrow 6 \times 5 = \left( {n - 4} \right) \times \left( {n - 5} \right)$
On multiplication, we get,
$\Rightarrow 30 = {n^2} - 4n - 5n + 20$
On rearranging, we get,
$\Rightarrow {n^2} - 9n + 20 - 30 = 0$
On simplifying we get,
$\Rightarrow {n^2} - 9n - 10 = 0$
Now we can solve the equation by factorisation.
For that we can split the middle term.
$\Rightarrow {n^2} - 10n + n - 10 = 0$
On taking the common factors from the 1st 2 and last 2 terms, we get,
$\Rightarrow n\left( {n - 10} \right) + \left( {n - 10} \right) = 0$
Again, by taking the common factors, we get,
$\Rightarrow \left( {n + 1} \right)\left( {n - 10} \right) = 0$
So, we can say that either $\left( {n + 1} \right) = 0$ or $\left( {n - 10} \right) = 0$
When $\left( {n + 1} \right) = 0$ ,
$\Rightarrow n = - 1$
This is not possible as n cannot be negative.
When $\left( {n - 10} \right) = 0$ ,
$\Rightarrow n = 10$
Now we can substitute the value of n in ${}^{12}{C_n}$
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12!}}{{10!\left( {12 - 10} \right)!}}$
On simplification we get,
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12!}}{{10!\left( 2 \right)!}}$
We know that $n! = n \times \left( {n - 1} \right)!$ . On using this relation, we get
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12 \times 11 \times 10!}}{{10! \times 2}}$
On cancelling the common terms, we get,
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12 \times 11}}{2}$
On simplification, we get,
$\Rightarrow {}^{12}{C_{10}} = 66$

So, the required value is 66.

Note:
Alternative method to solve this problem is given by,
We are given that ${}^n{C_4} = {}^n{C_6}$ .
We know that ${}^n{C_r} = {}^n{C_{n - r}}$ . On applying this on the RHS, we get,
$\Rightarrow {}^n{C_4} = {}^n{C_{n - 6}}$
Now we can equate the subscripts of the combination,
$\Rightarrow 4 = n - 6$
On rearranging, we get,
$\Rightarrow n = 4 + 6$
On simplification we get,
$\Rightarrow n = 10$
Now we can substitute the value of n in ${}^{12}{C_n}$
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12!}}{{10!\left( {12 - 10} \right)!}}$
On simplification we get,
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12!}}{{10!\left( 2 \right)!}}$
We know that $n! = n \times \left( {n - 1} \right)!$ . On using this relation, we get
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12 \times 11 \times 10!}}{{10! \times 2}}$
On cancelling the common terms, we get,
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12 \times 11}}{2}$
On simplification, we get,
$\Rightarrow {}^{12}{C_{10}} = 66$
So, the required value is 66.