Question

Find n if (i) 3 (n+1P4) = NP5,​

Answer 1

Answer:

$\red { Value \: of \: n } \green {= 9 }$

Step-by-step explanation:

$Given \: 3\left( ^{n+1}P_{4} \right) = ^{n}P_{5}$

$\implies 3 \left( \frac{(n+1)!}{(n+1-4)!}\right) = \frac{n!}{(n-5)! }$

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

$\implies 3 \left( \frac{(n+1)!}{(n-3)!}\right) = \frac{n!}{(n-5)! }$

$\implies 3 \left( \frac{(n+1)\cdotn!}{(n-4)\cdot (n-4)\cdot(n-5)!}\right) = \frac{n!}{(n-5)! }$

$\implies 3 \left( \frac{n+1}{(n-3)(n-4)}\right) =1$

$\implies 3(n+1) = (n-3)(n-4)$

$\implies 3n + 3 = n^{2} - 7n + 12$

$\implies 0 = n^{2} - 7n + 12 - 3n - 3$

$\implies n^{2} - 10n + 9 = 0$

/* Splitting the middle term, we get

$\implies n^{2} - 9n - 1n + 9 = 0$

$\implies n( n - 9 ) - 1( n - 9 ) = 0$

$\implies (n-9)(n-1) = 0$

$\implies n - 9 = 0 \: Or \: n - 1 = 0$

$\implies n = 9 \: Or \: n = 1$

$n = 1 \: is \: not \: possible$

Therefore.,

$\red { Value \: of \: n } \green {= 9 }$

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