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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 5 of 10.
Srinivas said:   1 decade ago
Can any one explain how we apply formula ncr = nc(n-r) to only first and last term only?
Atul said:   1 decade ago
I am not satisfied from the above ways because I don want to use simpler method.

Tell me why did we apply formula only on 1st and last terms. Why not on other terms?
Galla said:   1 decade ago
Wow your various explanations really did help.
Sarasa said:   10 years ago
Why didn't nC (n-r) method use for 6c2? Anyone please explained it?

I didn't get answer for this.
Nikhil said:   10 years ago
Actually one need not use nCr=nC (n-r) concept. Just use the concept that nCr= n!/(r!* (n-r) !). You will get the answer.
GunaVel said:   10 years ago
Any another easy method?
Payal said:   10 years ago
Second step is not clear.
Rahul said:   10 years ago
Why we are not using nc(n-r) formula in all in 2nd step?
Manek said:   10 years ago
I have understood this sum.
Shubham said:   10 years ago
Guys just calculate the opposite of that, case where no boy is there.

4c4 = 1.

Then all case which are (4+6) c4 = 210.

Then answer is 210-1 = 209.
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