Answer with Explanation:

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 x 4 / 3 x 2 x 1) + (6 x 5 / 2 x 1)   
 
 = (24 + 90 + 80 + 15) 
 
 = 209.

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