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 1 of 6.
Amanat said:
4 years ago
Thank you so much @Satish.
(1)
Saideep Uikey said:
6 years ago
Hi Folks, In below link, look at the formula in point no. 6, note no. ii
https://www.indiabix.com/aptitude/permutation-and-combination/formulas
With the formula mentioned there, that they have written as (7C5 x 3C2) = (7C2 x 3C1). Even I was confused but when I saw that formula in above link then I understood the concept very clear. School student won't face difficulty while understanding the solution because they must be knowing this formula already but people who are out of touch with math for lot of year like me will take time to digest these solutions. Hope my comment helps everyone. Thanks.
https://www.indiabix.com/aptitude/permutation-and-combination/formulas
With the formula mentioned there, that they have written as (7C5 x 3C2) = (7C2 x 3C1). Even I was confused but when I saw that formula in above link then I understood the concept very clear. School student won't face difficulty while understanding the solution because they must be knowing this formula already but people who are out of touch with math for lot of year like me will take time to digest these solutions. Hope my comment helps everyone. Thanks.
(4)
Kamal said:
7 years ago
Anybody help to solve this problem?
7 members have to be selected from 12 men and 3 women such that no two women can come together in how many ways we can select them?
7 members have to be selected from 12 men and 3 women such that no two women can come together in how many ways we can select them?
(1)
Keerthana said:
7 years ago
The answer is 63. I agree.
(2)
Ashutosh Sahoo said:
7 years ago
5 out of 7 and 2 out of 3.
so,
=7c5 x 3c2.
= (7! / 5! x 2!) x (3! / 2! x 1!),
= (7 x 6 / 2) x 3,
=( 42 / 2 ) x 3,
= 21 x 3,
= 63 (ans).
so,
=7c5 x 3c2.
= (7! / 5! x 2!) x (3! / 2! x 1!),
= (7 x 6 / 2) x 3,
=( 42 / 2 ) x 3,
= 21 x 3,
= 63 (ans).
(3)
Mike said:
7 years ago
Where 6 comes from? Please explain me.
Shree said:
7 years ago
Thanks for giving an explanation.
(2)
Bhavani said:
7 years ago
How will get 7c2* 3c1?
Please explain me.
Please explain me.
Kyosi billy said:
8 years ago
Thanks for explaining it.
Thatchayini said:
8 years ago
In that question, they didn't mention anything about permutation or combination. so we can try it in both ways.
7C5 * 3C2 = (7 * 6 * 5 * 4 * 3)/(5 * 4 * 3 * 2 * 1) * (3 * 2)/(2 * 1).
= 7 * 3 * 3 = 21 * 3 = 63.
7C5 * 3C2 = (7 * 6 * 5 * 4 * 3)/(5 * 4 * 3 * 2 * 1) * (3 * 2)/(2 * 1).
= 7 * 3 * 3 = 21 * 3 = 63.
(4)
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