Discussion :: Permutation and Combination
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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?
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Answer : Option D
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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