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WAEC Further Mathematics 2013 Paper

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Question : 39

Total: 40

If

\frac{{}^nC_3}{{}^nP_2} = 1

, find the value of n.

8
7
6
5

Solution:

Concept:

Use the standard formulas for combinations and permutations and simplify the ratio.

Explanation:

{}^nC_3 = \frac{n!}{3!(n-3)!}

{}^nP_2 = \frac{n!}{(n-2)!}

So,

\frac{{}^nC_3}{{}^nP_2} = \frac{n!}{3!(n-3)!} \div \frac{n!}{(n-2)!}

This equals

\frac{n!}{3!(n-3)!} \times \frac{(n-2)!}{n!}

Cancel

n!

to get

\frac{(n-2)!}{3!(n-3)!}

Since

(n-2)! = (n-2)(n-3)!

, we have

\frac{(n-2)(n-3)!}{3!(n-3)!} = \frac{n-2}{6}

Given

\frac{n-2}{6} = 1

, so

n-2 = 6

Hence,

n = 8

Answer:

n = 8

, which is option A.

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