# From a class of 12 boys and 10 girls, 10 students are to be chosen for the competition, at least including 4 boys and 4 girls. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made?

Given that 10 students need to be selected from 12 boys and 10 girls including at least 4 boys and 4 girls.

It is also told that the two girls must be includes who won prizes last year.

The cases that satisfy these conditions are:

i. Selecting 6 boys and 4 girls (in which 2 girls are already included)

ii. Selecting 5 boys and 5 girls (in which 2 girls are already included)

iii. Selecting 4 boys and 6 girls. (in which 2 girls are already included)

Let us assume the total no. of ways of selection be N,

N = (no. of ways of selecting 6 boys and 2 girls from remaining 12 boys and 8 girls) + (no. of ways of selecting 5 boys and 3 girls from remaining 12 boys and 8 girls) + (no. of ways of selecting 4 boys and 4 girls from remaining 12 boys and 8 girls)

Since, two girls are already selected,

N = (12C6 × 8C2) + (12C5 × 8C3) + (12C4 × 8C4)

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

N = (924 × 28) + (792 × 56) + (495 × 70)

N = 25872 + 44352 + 34650

N = 104874

The total number of ways of product is 11 ways.

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