Question
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?
Answer: Option D
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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)
= \(\left(6\times4\right)+\left(\frac{6\times5}{2\times1}\right)+\left(\frac{6\times5\times4}{3\times2\times1}\right)+\left(\frac{4\times6}{2\times1}\times4\right) +\left(\frac{6\times5}{2\times1}\right)\)
= (24 + 90 + 80 + 15)
= 209.
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