Question

If ⁿP₃ + ⁿC_n-2 = 14n, then the value of n is

• A. 5
• B. 6
• C. 8
• D. 10

Answer 1

The correct option is A

5

Step(i) :- General Formula of ${}^{n}p_r$

${}^{n}p_r=\frac{n!}{(n-r)!}$

Step(ii) :- Explanation

${}^{n}p_3+{}^{n}C_{n-2}=14n$

$$\begin{gathered} \left[\frac{n!}{(n-3)!}\right]+\left[\frac{n!}{(n-2)!\times 2!}\right]=14n \\ n(n-1)(n-2)+\frac{n(n-1)}{2}=14n \\ n(n-1)\left[n-2+\frac{1}{2}\right]=14n \\ n(n-1)(2n-3)=28n \\ (n-1)(2n-3)=28 \\ 2n^2-5n-25=0 \\ (n-5)(2n+5)=0 \\ n=5or\frac{-5}{2} \end{gathered}$$

As n cannot be negative.

Hence, the value of n is 5.

Hence, Option (a) is the correct answer