Question

Find the value of n such that nP2=5* nP3

Answer 1

$Given \: ^{n}P_{2} = 5 \times ^{n}P_{3}$

$\implies \frac{n!}{(n-2)!} = 5 \times \frac{n!}{(n-3)!}$

$\boxed { \pink { ^{n}P_{r} = \frac{n!}{(n-r)!} }}$

$\implies \frac{n!}{(n-2)(n-3)!} = 5 \times \frac{n!}{(n-3)!}$

$\implies \frac{1}{(n-2)} = 5 \times 1$

$\implies n-2 = \frac{1}{5}$

$\implies n = \frac{1}{5} + 2$

$\implies n = \frac{1+ 10}{5}$

$\implies n = \frac{11}{5}$

Therefore.,

$\red {Value \:of \:n } \green {= \frac{11}{5} }$

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