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

Answer:

c is the answer

Step-by-step explanation:

Answer 2

The number of permutations of $m(0\leq m\leq n)$ objects from $n$ objects is $P(m,n)=\frac{n!}{(n-m)!}$.

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

Correct choice is 120.