Given #10;that: #10; #10; #10; #10; ^np_4=20×^np_2 #10; #10; n!(n 4)!=20× #10;n!(n 2)! #10; #10; (n 2)(n 3)(n 4)!=20× (n 4)! ...

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Fractional Exponent

nP4=20× nP2 f...

Question

$^{n} P_{4} = 20 \times^{n} P_{2}$ find $n$

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Solution

Given that:

$^{n} p_{4} = 20 \times^{n} p_{2}$

$\frac{n!}{(n - 4)!} = \frac{20 \times n!}{(n - 2)!}$

$(n - 2) (n - 3) (n - 4)! = 20 \times (n - 4)!$

$n^{2} - 5 n + 6 = 20$

$n^{2} - 5 n - 14 = 0$

$n^{2} - 7 n + 2 n - 14 = 0$

$(n - 7) (n + 2) = 0$

$n = 7, - 2 (N o t c o n s i d e r a b l e)$

Hence, $n = 7$

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