Let S be the set of five digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?
Answer: B
Let O and E represent odd and even digits respectively.
∴ S can have digits of the form:
O _ O _ E or O _ E _ O or E _ O _ O
Case 1: O _ O _ E
The first digit can be chosen in 3 ways out of 1, 3 and 5
The third can be chosen in 2 ways.
The fifth digit can be chosen in 2 ways after which the second and fourth digits can be chosen in 2 ways.
∴ There are 3 × 2 × 2 × 2 = 24 ways in which this number can be written.
12 out of these ways will have 2 in the rightmost position and 12 will have 4 in the rightmost position.
∴ The sum of the rightmost digits in Case 1 = (12 × 2) + (12 × 4) = 72
Case 2: O _ E _ O
This number can again be written in 24 ways.
8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the rightmost position and 8 will have 5 in the rightmost position.
Thus the sum of the rightmost digits in Case 2 = (8 × 1) + (8 × 3) + (8 × 5) = 72
Case 3: E _ O _ O
This number can also be written in 24 ways.
As in Case 2, 8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the rightmost position and 8 will have 5 in the rightmost position.
∴ The sum of the rightmost digits in Case 3 = (8 × 1) + (8 × 3) + (8 × 5) = 72
∴ The sum of the digits in the rightmost position of the numbers in S = 72 + 72 + 72 = 216
Hence, option (b).