Question

Find the number of positive three-digit integers whose digits are among 9, 8, 7, 5, 3, 1 in which none of the digits are the same.

15
120
216

Answer 1

c is the answer

Explanation:

Answer 2

The number of permutations of
`m(0\leq m\leq n)` objects from
`n` objects is
`P(m,n)=(n!)/((n-m)!)`.

The number of ways of choosing 3 digits from 6 digits without replacement ( order is important) is
`P(6,3)=(6!)/((6-3)!) =(6!)/(3!) =4*5*6=120`. This is the number of all possible permutations of choosing 3 from 6.

Correct choice is 120.