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Permutation

If nPr=990 ...

Question

If ${n_{P}}_{r} = 990$ and ${n_{C}}_{r} = 165$ , then find the value of r.

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Solution

${n_{P}}_{r} = r! . {n_{C}}_{r}$
$r! = \frac{990}{165}$

$r! = 6$
$r = 3$

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If $^{n} P_{r} = 720$ and $^{n} C_{r} = 120$ , find the value of r.

^{n} P_{r} =^{n} P_{r + 1}

^{n} C_{r} =^{n} C_{r - 1}

, then the value of

r

^{n} P_{r} =^{n} P_{r + 1}

^{n} C_{r} =^{n} C_{r - 1}

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^{n} C_{r} =^{n} C_{r}_{-}_{1}

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^{n} P_{r} = 360

^{n} C_{r} = 15

then find the value of r

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