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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 9 of 10.
Jessie said:   1 decade ago
Udaya, your answer was really simpler.
Kavya said:   9 years ago
Why there is no 4C4 means 4 girls?
Alok said:   9 years ago
Can it not be written as 6 * 9C3?
Manek said:   10 years ago
I have understood this sum.
Chiru said:   9 years ago
Its easy.

10c4 - 4c4 =209.
Imnikesh said:   7 years ago
How 6c4 is reduced to 6c2?
Hmy nj said:   9 years ago
Why is it wrong 6C1 * 9C3?
Payal said:   10 years ago
Second step is not clear.
GunaVel said:   10 years ago
Any another easy method?
Kaushik said:   1 decade ago
Thank you Aditya..!
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