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

The correct option is B 18

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 in $\frac{3!}{2!}=3$ ways. Hence the required number of ways $=6\times 3=18$