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

120!!!!

Explanation:

Answer 2

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`