Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 1)
1.
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
Answer: Option
Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
 Required number of ways |
= (7C3 x 6C2) + (7C4 x 6C1) + (7C5) | |||||||||||
|
||||||||||||
|
||||||||||||
| = (525 + 210 + 21) | ||||||||||||
| = 756. |
Discussion:
137 comments Page 11 of 14.
Melvin N said:
1 month ago
nCr = nC(n-r) --> formula.
7C3 = 7C(7-3) = 7C4 = 35.
7C3 = 7C(7-3) = 7C4 = 35.
(1)
Krishan said:
1 decade ago
Anyone could explain why we used 7c2 in second step?
Johar said:
4 years ago
Anyone, Please solve this problem clearly to get it.
(1)
Sumanth reddy said:
2 years ago
How it's changed from 7c4 to 7c3? Please explain me.
(22)
Ajaykumar said:
10 years ago
7C3 = 35, 7!/(7-3)!3! = 7!/4!3! = 7*6*5/3*2*1 = 35.
Akash said:
1 decade ago
Can anyone please explain what does C imply here?
Abhimanyu said:
9 years ago
756 WILL BE CORRECT.
35 * 15 + 35 * 6 + 21= 756.
35 * 15 + 35 * 6 + 21= 756.
Daniel said:
10 years ago
I am still have little problem with the solving.
Daniel said:
10 years ago
I am still have little problem with the solving.
Ishak said:
9 years ago
Remarked first line 7c5 second line 7c2. How?
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