Question

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?

Answer 1

Here total number of persons = 8

Number of persons used (no person is repeated) = 2

$\therefore$ Number of permutations = ${}^8{P}_2$

$=\frac{8!}{6!}=\frac{8\times 7\times 6!}{6!}=56.$