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 3 of 6.
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).
Sravani said:
10 years ago
How 3c2 = 3c1?
BGT said:
1 decade ago
How can't we use any other e example?
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.
Manju said:
1 decade ago
nCr = n!/(n-r)!r!
= 7C5*3C2.
= (7*6/2*1)*3 = 63.
= 7C5*3C2.
= (7*6/2*1)*3 = 63.
Vishesh said:
1 decade ago
Same here what is this c? I don't understand this.
Shital said:
1 decade ago
I didn't understand 7c2*3c1.
Satish said:
1 decade ago
Step 1: 7C5 = (7*6*5*4*3)/(5*4*3*2*1) [ 7C5=> here from 7 factorial for 5 counts and divide totally by 5 factorial].
= (7*6)/(2*1) [here group the combinations again].
= 7C2.
So 7C5=7C2 = (7*6)/2 =21 ---->1.
Step 2: similarly 3C2= (3*2)/(2*1) = 3/1= 3 ----> 2.
Multiply(Combination) (1) and (2) = Ans = 63.
[ ie.. nCr = n!/r! = nC(n-r)].
= (7*6)/(2*1) [here group the combinations again].
= 7C2.
So 7C5=7C2 = (7*6)/2 =21 ---->1.
Step 2: similarly 3C2= (3*2)/(2*1) = 3/1= 3 ----> 2.
Multiply(Combination) (1) and (2) = Ans = 63.
[ ie.. nCr = n!/r! = nC(n-r)].
(2)
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