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:
138 comments Page 4 of 14.
Calesto said:
9 years ago
I don't understand why 7C5, in which 5is total number of persons to be chosen where 3 are men and remaining two are obviously women.
Thus, the answer should be 735.
Thus, the answer should be 735.
Manuel said:
9 years ago
Help me knowing how to solve this kind of equation I'm doing my Student Individual Learning Plan.
I need to proof that I was correct so help me please.
I need to proof that I was correct so help me please.
Udoy said:
9 years ago
When do we add or multiply ?
Sneha said:
9 years ago
Any one know the another method of solving?
(1)
Tshewang chophel said:
9 years ago
How to get 7c3=35?
Abhijit said:
9 years ago
I tried to solve like this.
Out of 7 places first 3 places to be filled by men so.. 7*6*5 = 42*5 ways but order does not matter so removing redundancies (42*5)/3!.
Similarly Remaining two places can be filled in (10*9)/2!.
Final answer is same as 7C3 * 10C2.
Out of 7 places first 3 places to be filled by men so.. 7*6*5 = 42*5 ways but order does not matter so removing redundancies (42*5)/3!.
Similarly Remaining two places can be filled in (10*9)/2!.
Final answer is same as 7C3 * 10C2.
Riya said:
9 years ago
I did not understand how we got 21 for 7C5?
(1)
Naina said:
9 years ago
What are the number of ways to select 3 men and 2 women such that one man and one woman are always selected?
The Answer is 30 and solution is 4c2 * 5c1.
But, I don't think it is right according to the condition given in the question.
Someone help me.
The Answer is 30 and solution is 4c2 * 5c1.
But, I don't think it is right according to the condition given in the question.
Someone help me.
Aryabhatt said:
9 years ago
Why don't we apply permutation after getting the answer 756 to get the number of ways those 5 persons can be arranged?
Felix said:
9 years ago
How is 7c3 = 35?
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