First slide

Combinations

Question

If $^{n} P_{r} = 1680$ and $^{n} C_{r} = 70$ then $n$ is equal to

Moderate

A

5
B

7
C

8
D

10

Solution

$\begin{array}{l} ^{n} P_{r} = (r!) (^{n} C_{r}) \Rightarrow 1680 = (r!) (70) \\ \Rightarrow r! = 24 \Rightarrow r = 4 \end{array}$
Also $1680 =^{n} P_{4} = n (n - 1) (n - 2) (n - 3)$
$\begin{array}{l} \Rightarrow (n^{2} - 3 n) (n^{2} - 3 n + 2) = 1680 \\ \Rightarrow {(n^{2} - 3 n + 1)}^{2} = 41^{2} \Rightarrow n^{2} - 3 n + 1 = 41 \\ \Rightarrow n^{2} - 3 n - 40 = 0 \Rightarrow (n + 5) (n - 8) = 0 \\ \Rightarrow n = 8 . \end{array}$

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