First slide

Combinations

Question

If $^{n} P_{r} = 2520 {and}^{n} C_{r} = 21$ then $n$ is is equal to:

Moderate

A

6
B

7
C

8
D

10

Solution

$(r!) (^{n} C_{r}) =^{n} P_{r} \Rightarrow (r!) (21) = 2520$
$\Rightarrow r! = 120 \Rightarrow r = 5$
Now $^{n} C_{5} = 21 = \frac{1}{2!} (7 \times 6) = \frac{7!}{5! 2!}$
$\Rightarrow n = 7$

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