Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 11)
11.
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
Answer: Option
Explanation:
| Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = |  |
7 x 6 | x 3 |  |
= 63. |
| 2 x 1 |
Discussion:
51 comments Page 2 of 6.
Sravani said:
10 years ago
How 3c2 = 3c1?
Sam said:
9 years ago
Why we are taking as 7c2 * 3c1? Please explain it.
Anoushka said:
9 years ago
Sir I haven't understood why have we multiplied 3 with 7*6/2*1 *3.
SHIVAM SHUKLA said:
10 years ago
I didn't understand this problem. Please each and every step explain it.
Kunal said:
10 years ago
But why they did 7c5*3c2 as 7c2*3c1?
Why they directly not calculate 7c5*3c1?
Why they directly not calculate 7c5*3c1?
Harish julai said:
10 years ago
Hello @Sravani.
If n>r, then ncr becomes nc(n-r).
If n>r, then ncr becomes nc(n-r).
BOND81 said:
1 decade ago
According to me the answer should be 126 as:
If we consider men in 1 group that is 7c5 and then women in 1 group that is 3c2 we can arrange these two groups in 2! ways.
So answer should be 7c5*3c2*2! = 126.
If we consider men in 1 group that is 7c5 and then women in 1 group that is 3c2 we can arrange these two groups in 2! ways.
So answer should be 7c5*3c2*2! = 126.
Mayank said:
9 years ago
@Sudha.
As know this formula n!/r!(n-r)!.
7!/5!2!
3!/2!1!
The answer will be same .
What they have done is try to reduce one step.
As know this formula n!/r!(n-r)!.
7!/5!2!
3!/2!1!
The answer will be same .
What they have done is try to reduce one step.
Mariyam said:
9 years ago
@Satish your explanation is very understandable.
Dhayithri said:
9 years ago
Here we can also taken as 3c2 as 3c1 because factorial of those two is equal.
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