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

Question

asked Jun 19, 2018 74.2k views

2 Answers

Dave Cadwallader · Answer 1 · 2018-06-23T16:45:36+0000

Answer: 120

Explanation:

The given digits : 9, 8, 7, 5, 3, 1

Number of digits : 6

If repetition of digit is not allowed then we use permutations.

We know that the number of permutations of n things taking r at a time is given by :-

^nP_r=(n!)/((n-r)!)

Now, the number of permutations of 6 digits taking 3 at a time is given by :-

$^6P_3=(6!)/((6-3)!)\\\\=(6*5*4*3!)/(3!)=120$