From a committee of $8$ persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

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Solution

Number of People $= n = 8$
Number of people to be chosen for $2$ positions $= r = 2$
$∴$ Number of ways we can choose a chairman and a vice-chairman
$=^{8} P_{2} = \frac{8!}{(8 - 2)!} = \frac{8!}{6!} = 8 \times 7$
$= 56 ways$ .

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