Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 6)
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?
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 numberof ways |
= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) | |||||||||||||||||||
| = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) | ||||||||||||||||||||
|
||||||||||||||||||||
| = (24 + 90 + 80 + 15) | ||||||||||||||||||||
| = 209. |
Discussion:
92 comments Page 5 of 10.
Kashika shekhar said:
1 decade ago
I am not able to apply different types of logic in different questions. Please tell me some logic for appying the formulas.
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.
Sam said:
1 decade ago
Is there anyone to explain why this particular question cannot be solved in the same method which were solved above?
Shilpa said:
1 decade ago
6c2= 6!/2!(6-2)!
= 6!/2!*4!
= 6*5*4! 2*4!
Then 4! will be cancel.
= 6*5/2.
By solving it.
= 15 got in result.
= 6!/2!*4!
= 6*5*4! 2*4!
Then 4! will be cancel.
= 6*5/2.
By solving it.
= 15 got in result.
Aditi Muley said:
1 decade ago
In third step last 1 is 6C2, I am confused in the fourth step about how we solved 6C2 that we got 15 in result.
Nithin said:
5 years ago
Why can't we do 6C1*9C3?
First choose a boy from group of 6 and then choose any three from remaining 9 people.
First choose a boy from group of 6 and then choose any three from remaining 9 people.
Prajwal BK said:
2 years ago
1 case is missed out i.e., 4 girls can also be one more case.
So that makes up to 5 possible scenarios.
So that makes up to 5 possible scenarios.
(5)
Pri said:
1 decade ago
I'm a bit confused.
In 1st step : ... + (6c1 x 4c3) + ...
In 2nd step : ... + (6c1 x 4c1) + ... How ?
In 1st step : ... + (6c1 x 4c3) + ...
In 2nd step : ... + (6c1 x 4c1) + ... How ?
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.
I didn't get answer for this.
Pallavi said:
1 decade ago
Another easy method is
1-(4C4/10C4)
=1-(1/210)
=209/210
Therefore 209 times out of possible 210.
1-(4C4/10C4)
=1-(1/210)
=209/210
Therefore 209 times out of possible 210.
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