Question

Find the value of n such that ${}^n{p}_5=42{(}^n{p}_3):n>4$

Answer 1

${n_p}_5=42\left({n_p}_3\right)$

$\frac{n!}{(n-5)!}=42\left(\frac{n!}{(n-3)!}\right)$

$(n-3)!=42(n-5)!$

$(n-3)(n-4)=42$

$n^2-7n-30=0$

$n^2-10n+3n-30=0$

$n(n-10)+3(n-10)=0$

$(n-10)(n+3)=0$

Therefore, $n=10$