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

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Solution

From a committee of $8$ person, a chairman and a vice chairman are to be chosen in such a way that one person cannot hold more than one position.
Here, the number of ways of choosing a chairman is the permutation of $8$ different objects taken $2$ at a time.
Thus, required number of ways= $^{8} P_{2} = \frac{8!}{(8 - 2)!} = \frac{8!}{6!} = \frac{8 \times 7 \times 6!}{6!} = 8 \times 7 = 56$

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