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