Question

${}^{n-1}{P}_3:{}^n{P}_4=1.4$ then what is the value of $n$.

Answer 1

Given,

$\frac{{}^{n-1}{p}_3}{{}^n{p}_4}=\frac{1}{4}$

$\frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}}=\frac{1}{4}$

$\frac{\frac{(n-1)!}{(n-4)!}}{\frac{n(n-1)!}{(n-4)!}}=\frac{1}{4}$

$\frac{1}{n}=\frac{1}{4}$

$n=4$

Hence, this is the answer.