Correct Answer: Option D
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
= (6 x 4) + {(6 x 5)/( 2 x 1) x (4 x 3)/( 2 x 1)} + [{(6 x 5 x 4)/(3 x 2 x 1)}x 4] + (6 x 5)/( 2 x 1)
= (24 + 90 + 80 + 15)
= 209.
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