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Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 6)
Permutation and Combination
6.
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?
159
194
205
209
None of these
Answer: Option
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 x 4 x 3 + 6 x 5 x 4 x 4 + 6 x 5
2 x 1 2 x 1 3 x 2 x 1 2 x 1
= (24 + 90 + 80 + 15)
= 209.

Discussion:
92 comments Page 10 of 10.
Aditya said:   1 decade ago
Hi Pri,
Its because nCr = nC(n-r). So 4C3 = 4C(4-3) = 4C1.
Its one of the formulaes.
Pri said:   1 decade ago
I'm a bit confused.

In 1st step : ... + (6c1 x 4c3) + ...
In 2nd step : ... + (6c1 x 4c1) + ... How ?
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