Aptitude - Permutation and Combination - Discussion
Discussion Forum : Permutation and Combination - General Questions (Q.No. 8)
8.
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
Answer: Option
Explanation:
| Required number of ways | = (8C5 x 10C6) | |||||||
| = (8C3 x 10C4) | ||||||||
|
||||||||
| = 11760. |
Discussion:
61 comments Page 5 of 7.
Charan said:
8 years ago
8c3 = 8!/3!*(8-3)! = 8!/3!*5! which is equal to 8c5 = 8!/5!*(8-5)! = 8!/5!*3!.
Charan said:
8 years ago
6men and 5 women can also be changed so it can be 11!*11760.
Gayudhaya said:
8 years ago
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
By using above formula just explain it.
By using above formula just explain it.
Tejal said:
8 years ago
Don't we have to divide (8C5*10C6) by total no of selection that is 18C11?
Nancy said:
8 years ago
Please help me to understand why the deduction (the second step) is done?
Pankaj kumar said:
7 years ago
The second step involves the formula of nCr,
which is;
nCr = n!/(n-r)!r!.
which is;
nCr = n!/(n-r)!r!.
B S said:
7 years ago
The Answer should be 266.
Shivani Bairagi said:
7 years ago
Why we take 8C3 * 10C4?
I am not understanding. So please explain me in detail.
I am not understanding. So please explain me in detail.
Rahul said:
7 years ago
I am not getting this answer. Please, anyone, help me to get this.
Mohan said:
6 years ago
8C5 = (8*7*6*5*4)/(5*4*3*2*1).
= (8*7*6)/(3*2*1).
= 8C3.
10C6 = (10*9*8*7*6*5)/(6*5*4*3*2*1).
= (10*9*8*7)/(4*3*2*1),
= 10C4.
= (8*7*6)/(3*2*1).
= 8C3.
10C6 = (10*9*8*7*6*5)/(6*5*4*3*2*1).
= (10*9*8*7)/(4*3*2*1),
= 10C4.
(5)
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