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 8 of 10.
Hitesh said:
1 decade ago
What about when there was asked for at most 2 boys?
Shreu said:
1 decade ago
But the also there the children how it is possible?
Shanmuga kumar said:
1 decade ago
How do you get 4 members in group? please explain.
Shivaraj said:
9 years ago
@Neema.
It's mentioned, read question clearly.
It's mentioned, read question clearly.
Galla said:
1 decade ago
Wow your various explanations really did help.
Sugumar said:
10 years ago
Why didn't you take 4 girls out of 4 girls?
Dhanraj said:
6 years ago
Thanks for explaining it in detail @Shivam.
P.krishna reddy said:
5 years ago
Atleast one = Total - None = 10c4-1 = 209.
(4)
Abhishek said:
1 decade ago
Udaya its easy. But its more calculative.
Sathvika said:
6 years ago
4c2 = 4!/2!*(4-2)!.
= 4 * 3/2 * 2,
= 3.
= 4 * 3/2 * 2,
= 3.
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