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 12 of 14.
Siva kumar said:
9 years ago
Nice solution @Nagu.
Ishak said:
9 years ago
Remarked first line 7c5 second line 7c2. How?
Abhimanyu said:
9 years ago
756 WILL BE CORRECT.
35 * 15 + 35 * 6 + 21= 756.
35 * 15 + 35 * 6 + 21= 756.
Rutuja said:
9 years ago
Someone, please give the solution.
In a group of 6 people, there are 3 Indians and 3 Chinese. How many subsets can be created such that there are at least 1 Indian in each subset?
In a group of 6 people, there are 3 Indians and 3 Chinese. How many subsets can be created such that there are at least 1 Indian in each subset?
Harsh said:
8 years ago
It's correct that we select 3 men from 7 and 2 women from 6. But why do we need to multiply it to get total no of ways?
Shoeib said:
8 years ago
Yeah, I too agree @Harsh.
Ali said:
8 years ago
@Harsh.
I think we need to multiply so that we get the total number of ways we can choose 3 men from a total of 13 men and women.
I think we need to multiply so that we get the total number of ways we can choose 3 men from a total of 13 men and women.
Mamta Patel said:
8 years ago
The total number of combinations possible: 13C5=1287.
Then Calculate the number of combinations with less than 3 men in them: 7C2*6C3=21*20=420 7C1*6C4=7*15=105 7C0*6C5=1*6=6.
After that deduct the number of combinations with less than 3 men from total number of combinations: 1287-(420+105+6)= 756.
Then Calculate the number of combinations with less than 3 men in them: 7C2*6C3=21*20=420 7C1*6C4=7*15=105 7C0*6C5=1*6=6.
After that deduct the number of combinations with less than 3 men from total number of combinations: 1287-(420+105+6)= 756.
Aryabhatt said:
8 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:
8 years ago
How is 7c3 = 35?
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