Question

$\bold {Find \: n \: if \: ^{n - 1} P_{3} : {}^{n} P_{4} = 1 :9}$


PLEASE ANSWER MY QUESTION​

Answer 1

$\bold{ \frac{^{n - 1} P_{3} }{{}^{n} P_{4} } = \frac{1}{9} }$

$\bold{\implies \: \frac{(n - 1)(n - 2)(n - 3)}{n(n - 1)(n - 2)(n - 4)} = \frac{1}{9}}$

$\bold{\implies \: \frac{1}{n} = \frac{1}{9}}$

Cross multiplying n=9

Answer 2

given★

• $\bold {Find \: n \: if \: ^{n - 1} P_{3} : {}^{n} P_{4} = 1 :9}$

find ★

• n value

EVALUTION★

• we know that

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

• $= > \frac{ \frac{(n - 1)!}{(n - 1 - 3)!} }{ \frac{n!}{(n - 4)!} } = \frac{1}{9} \\ \\ \\ = > \frac{ \frac{(n - 1)!}{(n - 4)!} }{ \frac{n!}{(n - 4)!} } = \frac{1}{9} \\ \\ \\ = > \frac{(n - 1)!}{n(n - 1)!} = \frac{1}{9} \\ \\ \\ = > \frac{1}{n} = \frac{1}{9} \\ \\ \\ = > n = 9$
• permutation formula

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I hope it helps you