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

216 is a positive three-digit interger

Answer 2

Answer: 120

Step-by-step 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 :-

$^nP_r=\dfrac{n!}{(n-r)!}$

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

$^6P_3=\dfrac{6!}{(6-3)!}\\\\=\dfrac{6\times5\times4\times3!}{3!}=120$