Question

Find the numbers that can be formed by using all digits 1,2,3,4,3,2,1 so that the odd digits always occupy the odd places

Answer 1

There are 4 odd places(1,3,3,1) and 3 even places(2,4,2).

Odd numbers can be arranged in 4!/2!2!.(two 1's and 3's)

Even numbers can be arranged in 3!/2!(two 2's)

Required numbers are = 4!/2!2! * 3!/2!

                                      = 6 * 3

                                      = 18.


Hope this helps!