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 10 of 10.
Divyesh Khunt said:
1 year ago
Another easy method here;
It's given at least 1 boy meaning (>=1) so we can subtract calculate 0 boys selected and subtract it from the total combinations possible.
Combination(boys>=1) =total combination - combination (0boys selected),
= 10C4 - 6C0 * 4C4.
= 210 -1.
= 209.
It's given at least 1 boy meaning (>=1) so we can subtract calculate 0 boys selected and subtract it from the total combinations possible.
Combination(boys>=1) =total combination - combination (0boys selected),
= 10C4 - 6C0 * 4C4.
= 210 -1.
= 209.
(20)
Gayathri m said:
1 week ago
The second step is not determined.
i.e (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
i.e (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2).
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