Question

The number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits always occupy odd places, is

• A. 24
• B. 18
• C. 12
• D. 30

Answer 1

Answer: (B)

18

Answer 2

The 4 odd digits 1, 3, 3, 1 can be arranged in the 4 odd places, in $\frac{4!}{2!2!} = 6$ ways and 3 even digits 2, 4, 2 can be arranged in the three even places $\frac{3!}{2!} = 3$ways. Hence the required number of ways = 6 × 3 = 18.