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) | |||||||||||
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||||||||||||
|
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| = (525 + 210 + 21) | ||||||||||||
| = 756. |
Discussion:
137 comments Page 3 of 14.
Chithra said:
1 decade ago
Combination is nothing but we are going to select from a particular group. In permutation also we will select from group, after selecting we need to arrange it according to the given condition.
i.e.) Combination -> selection.
Permutation -> selection followed by arrangement.
i.e.) Combination -> selection.
Permutation -> selection followed by arrangement.
Abhijit said:
8 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.
Naina said:
8 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.
Nikhil said:
1 decade ago
Why would the logic 7C3 * 10C2 not work? First we've choosen 3 Men - 7C3. Now we have 10 people left (which included both men and women). Then we need to select any 2 people either men or women so 10C2. Please Clarify.
Vishwanath said:
1 year ago
We need a minimum 3 men, so the maximum can be all 5 men in the committee.
3 men AND 2 women = 7C3 X 6C2 = 525.
4 men AND 1 women =7C4 X 6C1 = 210.
all 5 men = 7C5 = 21.
Add them 525 + 210 + 21 = 756.
3 men AND 2 women = 7C3 X 6C2 = 525.
4 men AND 1 women =7C4 X 6C1 = 210.
all 5 men = 7C5 = 21.
Add them 525 + 210 + 21 = 756.
(25)
Kiprono Langat Esau said:
9 years ago
Kindly, solve this.
The committee of six is to be formed from a group of seven engineers and four mathematicians, how many different committees can be formed if at least two mathematicians are always to be included?
The committee of six is to be formed from a group of seven engineers and four mathematicians, how many different committees can be formed if at least two mathematicians are always to be included?
Prashant Olekar said:
1 decade ago
According to combination rule we can write 7c3 as (7*6*5/3*2*1).
Similarly.
= (7c3*6c2) + (7c4*6c1) + (7c5).
= ((7*6*5/3*3*1)*(6*5/2*1)) + ((7*6*5*4/4*3*2*1)*(6)) + ((7*6*5*4*3/5*4*3*2*1)).
= 525+210+21.
= 756.
Similarly.
= (7c3*6c2) + (7c4*6c1) + (7c5).
= ((7*6*5/3*3*1)*(6*5/2*1)) + ((7*6*5*4/4*3*2*1)*(6)) + ((7*6*5*4*3/5*4*3*2*1)).
= 525+210+21.
= 756.
Dagwidhi said:
7 years ago
Why can't we do this? Select 3 out of 7 men. Because that's the main condition. Then select 2 women out of 6/1 man out of 4 and 1 woman out of 6/ 2 men out of 4.
i.e 7c3 (4c2 + 6c2 + 4c1. 6c1)?
Please tell me.
i.e 7c3 (4c2 + 6c2 + 4c1. 6c1)?
Please tell me.
Gokul said:
1 decade ago
It is simple to identify the given problem is permutation or combination. It is just by common sense ya. In permutation, there is arrangement of things. But in combination it is selection. I hope you understand.
Shiva said:
1 decade ago
@Olga.
You have removed only the condition where you have only two men, but you have to remove the condition where one men and no men are present. i.e. 7c2*6c3 + 7c1*6c4 +7c0*6c5 has to be removed from 13c5.
You have removed only the condition where you have only two men, but you have to remove the condition where one men and no men are present. i.e. 7c2*6c3 + 7c1*6c4 +7c0*6c5 has to be removed from 13c5.
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