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.