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?

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= ${}^8{P}_2=\frac{8!}{(8-2)!}=\frac{8!}{6!}=\frac{8\times 7\times 6!}{6!}=8\times 7=56$