We are asked to find in how many 3 digit numbers 2 and 6 are both present. So we can employ the following strategy:

1) select the third digit
2) arrange the 3 digits
3) remove invalid cases
4) remove repeated cases

1) no of ways of selecting the third digit (any one from 0 to 9)= 10C1 = 10
2) no. of ways of arrangement of 3 digits = 3!
So total arrangements = 10*3! = 60
3) invalid cases are the one where 0 is the first digit. There are only 2 such numbers ( 026 and 062 )
invalid cases = 2
4) cases are repeated when the third digit is either 2 or 6
For each of the digits, the repeated cases are 3!/2! = 3
Hence total repeat cases for the 2 digits = 2 * 3 = 6

Hence required number = 60-2-6 = 52.

Let me know if this helps

hey_thr67 wrote:How many different three-digit numbers contain both the digit 2 and the digit 6?

(A) 52
(B) 54
(C) 56
(D) 60
(E) 62

OA is A.

You could consider two mutually exlusive cases:

(1) three different digits, two of them being 2 and 6. There are 8 choices for the 3rd digit and 3! to arrange these 3 digits. However, 026 and 062 must be excluded TOTAL 48 - 2 = 46

(2) no digit other than 2 and 6. Decide which will be repeated (2 choices) and where the unique digit will go (3 choices) TOTAL 6

SUM 52