If $^{n} P_{r} = 5040 (^{n - 1} C_{r - 1} +^{n - 1} C_{r})$ then

r = 5

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r = 6

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r = 7

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r = 8

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Solution

The correct option is D $r = 7$
$^{n} P_{r} = 5040 (^{n - 1} C_{r - 1} +^{n - 1} C_{r})$
$^{n - 1} C_{r - 1} +^{n - 1} C_{r}$ $=^{(n - 1) + 1} C_{r - 1} +^{n - 1} C_{r} =^{n} C_{r}$
So,
$^{n} P_{r} = 5040^{n} C_{r}$
$⟹ \frac{n!}{(n - r)!} = 5040 \frac{n!}{r! (n - r)!}$
$⟹ r! = 5040 = 7!$
$⟹ r = 7$

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If $^{n} P_{r} = 5040 (^{n - 1} C_{r - 1} +^{n - 1} C_{r})$ ,then r =

If $^{n} P_{r} = 336,^{n} C_{r} = 56$ , then find n and r and hence find $^{n - 1} C_{r - 1}$

^{n + 1} C_{r + 1} :^{n} C_{r} :^{n - 1} C_{r - 1} = 11 : 6 : 3

n

r

n

r

1 \leq r \leq n

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

1 \leq r \leq n

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