Determine the number of permutations of 1 2 8 that fix exac

Determine the number of permutations of {1, 2, ..., 8} that fix exactly four elements. (Combinatorics)

Solution

The set {1, 2, ..., 8} has 8 elements. It is to be divided into 2 groups of 4 elements each.

If the 1^st number in the 1^st group is fixed, the 1^st number in the 2^nd group can be chosen in 7 different ways.

Once, the 1^st number in the 2 groups and the 2^nd number in the 1^st group are chosen, and the 2^nd numbers are chosen, the 2^nd number in the 2^nd group can be chosen in 5 different ways.

When the 1^st two numbers in both the groups and the 3^rd number in the 1^st group are chosen, the 3^rd number in the 2^nd group can be chosen in 3 different ways.

When the 1^st three numbers in both the groups and the 4^th number in the 1^st group are chosen, there is only 1 way to choose the 4^th number of the 2^nd group.

Hence, the total number of permutations in which the numbers {1, 2, ..., 8} can be fixed into 2 groups of 4 elements each is 7*5*3*1 = 105.