Aspire Faculty ID #13539 · Topic: JAMIA MILLIA ISLAMIA MCA 2024 · Just now
JAMIA MILLIA ISLAMIA MCA 2024

If $_nP_3 = 4 \times {_nP_2}$, find $n$.

Solution

${_nP_3} = \dfrac{n!}{(n - 3)!},\quad {_nP_2} = \dfrac{n!}{(n - 2)!}$ Given, $\dfrac{n!}{(n - 3)!} = 4 \times \dfrac{n!}{(n - 2)!}$ $\Rightarrow \dfrac{n(n - 1)(n - 2)!}{(n - 3)!} = 4 \times \dfrac{n!}{(n - 2)!}$ Simplify: $(n - 2) = 4 \Rightarrow n = 6$
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