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 5 of 14.
Y.Deepikamanoj said:
10 years ago
How many 3 digit numbers can be formed from 2, 3, 4, 5, 6, 7, 9, which are divisible by 5 and none of the digits is repeated? Please tell answer and method for this problem.
(1)
Anand kumar Gupta said:
6 years ago
Here 5 persons are select, and at least 3 men including this committee so we have
Men. Women
7c3 * 6c2 = 525
7c4 * 6c1 = 210
7c5 * 6c0 = 021
= 756 is the answer.
Men. Women
7c3 * 6c2 = 525
7c4 * 6c1 = 210
7c5 * 6c0 = 021
= 756 is the answer.
(1)
Suga said:
3 years ago
@All.
Here, some of the persons are confused in the second step because there was an incorrection where (7C4 x 6C1) + (7C5) were converted into + (7C3 x 6C1) + (7C2).
Here, some of the persons are confused in the second step because there was an incorrection where (7C4 x 6C1) + (7C5) were converted into + (7C3 x 6C1) + (7C2).
(14)
Calesto said:
8 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:
8 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.
Cyrus said:
1 decade ago
nCr = nCn-r
i.e n!/(r!*(n-r)!)
So, 7C4 = 7!/(4!*(7-4)!)
= 7*6*5*4*3*2*1/(4*3*2*1*(3*2*1))
= 7*6*5/3*2
= 35.
i.e n!/(r!*(n-r)!)
So, 7C4 = 7!/(4!*(7-4)!)
= 7*6*5*4*3*2*1/(4*3*2*1*(3*2*1))
= 7*6*5/3*2
= 35.
Anu said:
1 decade ago
5 ball are to be placed in 12 boxes, they are placed in three rows such that each row contain at least one ball in how many ways it can be placed.
Pankaj said:
1 decade ago
Would you please explain in question they ask only for the three man combination then why you have taken 4m and 1m and all five men possibilities?
Jay said:
4 years ago
@Nivetha.
We only have to find total combinations, not the probability. If we want to find probability then we have to divide it by 13C5.
We only have to find total combinations, not the probability. If we want to find probability then we have to divide it by 13C5.
(2)
Anshul Vyas said:
1 decade ago
Mr. Mohan is absolutely right.
We know that 7c3 is not equal to 7c1*6c1*5c1.
In the same way we can't write 10c2 for rest two members.
We know that 7c3 is not equal to 7c1*6c1*5c1.
In the same way we can't write 10c2 for rest two members.
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