Question

10. If P (n, 4 ) = 17 160 , then n what is the value of n? A. 9 B. 11 C. 13D. 14

Answer 1

This is a permutation problem.

The expression n permutation r is expressed as:

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

In like manner, n permutation 4 will be:

$\begin{gathered} P(n,4)=17160 \\ \frac{n!}{(n-4)!}=17160 \end{gathered}$

Evaluation the permutation operation above, we have:

$\begin{gathered} \frac{n!}{(n-4)!}=17160 \\ \frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}=17160 \\ (n-4)!\text{ cancels out (n-4)!, thus we have;} \\ n(n-1)(n-2)(n-3)=17160 \end{gathered}$

Expanding the Left hand side of the equation; we have:

$\begin{gathered} n^4-6n^3+11n^2-6n=17160 \\ n^4-6n^3+11n^2-6n-17160=0 \end{gathered}$

By factorization, the equation becomes;

$\begin{gathered} \mleft(n+10\mright)\mleft(n-13\mright)\mleft(n^2-3n+132\mright)=0 \\ (n^2-3n+132)\text{ is not factorizable and would also produce unreal roots, thus the value of n from the expression can't be correct} \\ n+10=0\text{ will produce n=-10, we can have a negative result for permutation problems} \\ \text{Thus, the correct answer is;} \\ n-13=0 \\ n=13 \end{gathered}$

Hence, the value of n is 13, option C