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WAEC Further Mathematics 2012 Paper

Question : 24

Total: 40

Given that

{}^{n}P_{r} = 90

{}^{n}C_{r} = 15

, find the value of r.

Solution:

$\; \frac{{}^{n}P_{r}}{{}^{n}C_{r}} = \frac{ \; \frac{n!}{(n-r)!} }{ \frac{n!}{(n-r)! r!} } = \; \frac{90}{15} = 6$

$\; \frac{n!}{(n-r)!} \times \; \frac{(n-r)! r!}{n!} = r! = 6$

$r = 3$

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