Question

If ${}^{n}{{C}_{n-2}}=15$then find n.

Answer 1

To solve this question at first we have to expand ${}^{n}{{C}_{n-2}}$ using a combination formula. Then we solve for the expression in terms of n and equate the expression with 15 to get an equation. By solving the obtained equation we will get the value of n.

Complete step-by-step answer:
From the definition of combination we know the formula,
${}^{n}{{C}_{r}}=\dfrac{n!}{(n-r)!r!}$ , .………………. (1), which shows number of ways to choose ‘r’ items from total ‘n’ items
By replacing ‘r’ with $\left( n-2 \right)$in eq. (1), we will get
{}^{n}{{C}_{n-2}}=\dfrac{n!}{\left\\{ n-\left( n-2 \right) \right\\}!\left( n-2 \right)!}
On solving bracket by adding n and - n, we get
${}^{n}{{C}_{n-2}}=\dfrac{n!}{2!\left( n-2 \right)!}$……………………………………………… (2)
We now that the factorial of a positive integer n is given by,
$n!=n\left( n-1 \right)\left( n-2 \right)\times ........\times 3\times 2\times 1$ ……………………………. (3)
For example,
$6!=6\times 5\times 4\times 3\times 2\times 1=720$
In a similar manner,
$\left( n-2 \right)!=\left( n-2 \right)\left( n-3 \right)\times ............\times 3\times 2\times 1$ ……………………… (4)
Now substituting the values of eq. (3) and (4) in eq. (2), we will get
$\Rightarrow {}^{n}{{C}_{n-2}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)\times ........\times 3\times 2\times 1}{2!\times \left( n-2 \right)\left( n-3 \right)\times ............\times 3\times 2\times 1}$
Eliminating all the common terms from denominator and numerator by cancelling them out, we get
$\Rightarrow {}^{n}{{C}_{n-2}}=\dfrac{n\left( n-1 \right)}{2}$………….. (5)
Here we got the expression for${}^{n}{{C}_{n-2}}$. To verify the expression let’s take some examples.
For$n=3$,

& {}^{3}{{C}_{3-2}}={}^{3}{{C}_{1}}=\dfrac{3!}{\left\\{ 3-1 \right\\}!1!}=\dfrac{3\times 2\times 1}{2\times 1\times 1}=3 \\\ & \dfrac{n\left( n-1 \right)}{2}=\dfrac{3\times 2}{2}=1 \\\ \end{aligned}$$ For$$n=5$$, $$\begin{aligned} & {}^{5}{{C}_{5-2}}={}^{5}{{C}_{3}}=\dfrac{5!}{\left\\{ 5-3 \right\\}!3!}=\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1\times 3\times 2\times 1}=10 \\\ & \dfrac{n\left( n-1 \right)}{2}=\dfrac{5\times 4}{2}=10 \\\ \end{aligned}$$ But according to question, $${}^{n}{{C}_{n-2}}=15$$ ............................................................ (6) Now comparing eq. (5) and (6) we will get, $$\begin{aligned} & \Rightarrow \dfrac{n\left( n-1 \right)}{2}=15 \\\ & \Rightarrow n\left( n-1 \right)=30 \\\ & \Rightarrow {{n}^{2}}-n-30=0 \\\ \end{aligned}$$ Factorizing, the quadratic equation by splitting, - n as -6n+5n, we get $$\begin{aligned} & \Rightarrow {{n}^{2}}-6n+5n-30=0 \\\ & \Rightarrow n(n-6)+5(n-6)=0 \\\ \end{aligned}$$ Taking, ( n -6 ) common from equation, we get $$\Rightarrow (n-6)(n+5)=0$$ On comparing Left hand side and right hand side, we get $$\Rightarrow n=6,-5$$…………………………………………. (7) But we know that n must be a positive integer from the definition of combination. Hence the required value of n is 6. **Note:** Factorial of a number n that is $$n!$$ defined only for any nonnegative integer n. Therefore here we cannot take the value for$$n=-5$$. Also, use can solve the quadratic equation $a{{x}^{2}}+bx+c=0$ to get value of x, by using quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Try not to make any calculation errors while solving the question, as it will change the final answer.