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### Discussion :: Permutation and Combination

1.  In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
2.  A. 159 B. 194 C. 205 D. 209 E. None of these

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) +$$[\frac { 6*5 } { 2*1 }$$x$$\frac { 4*3 } { 2 *1}$$]+[$$\frac { 6*5*4 } { 3*2*1 }$$x4]+$$[\frac { 6*5 } { 2*1 } ]$$

=(24 + 90 + 80 + 15)

= 209

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