Aptitude - Permutation and Combination - Discussion

Discussion :: 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?

[A]. 63
[B]. 90
[C]. 126
[D]. 45
[E]. 135

Answer: Option A

Explanation:

Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) = 7 x 6 x 3 = 63.
2 x 1


Priya said: (Jun 26, 2010)  
How? I can't understand this problem.

Sahithya said: (Sep 10, 2010)  
Hi priya,.

We have to select 5men out of 7 men, this can be done in 7C5 ways

= 7C2 [Ref: nCr=nC (n-r) ].

And also we hav to select 2women out of 3 women, this can be done in 3C2 ways = 3C1.

As these two combinations are compulsry, we hav to multiply 7C2 and 3C1.

I think you understand.

Binushree said: (Nov 25, 2010)  
How to multiply?

Karthik said: (Jan 7, 2011)  
Why can't we multiply 7C5 nd 3C2 directly ?

Suvarna said: (Feb 25, 2011)  
It takes more time to calculate 7c5 than 7c2& 3C2 than 3C1, to improve the speed we are doing like this.

Jat said: (Jun 4, 2011)  
(7C5 x 3C2) = (7C2 x 3C1)
This step i can't understand.
Please explain me.

Manasa said: (Jun 14, 2011)  
@jat
its just formula
n =n
c c
r n-r
n=7,r=5
7C5=7C2

Zalak said: (Jul 30, 2011)  
(7C5 x 3C2) = (7C2 x 3C1)

This step I can't understand. Please explain me.

Kavu said: (Sep 18, 2011)  
(7C2 x 3C1) = (7/2 x 6/1) x 3= 63.

How it comes?

Achutharaj said: (Sep 28, 2011)  
Basic formula nCr = nCn-r
so 7C7-5 = 7C2
3C3-2 = 3C1

Divya said: (Sep 29, 2011)  
How 3c2 change into 3c1?

Ankur Anand said: (Oct 8, 2011)  
Why combination why not permutation ?

Similar as that of the arranging of letters the people are diffrent so there can be many ways by permutation.

Nirmal said: (Nov 25, 2011)  
(7C2 x 3C1) = 7 x 6 x 3

How it happened explain please?

Teja said: (Dec 30, 2011)  
Hey see you said that 7c5*3c2 is long procedure. That the thing we take ncr formula. ?and second thing why we take that formula in this problem? am not?

Srikanth said: (Apr 26, 2012)  
@teja. As this a problem f selection you use combination. It means calculating 7c5 is a bit more tedious when compared to calculating 7c2.

Moni said: (Jun 27, 2012)  
For ex. Take 7C5, 5 is more than a half of 7. So we can use the formula nC (n-r) when are is more than half of n.

Similarly 3C2 is made as 3C1. I hope you have understood.

Niyati Bafna said: (Jul 27, 2013)  
Are there any more problems, perhaps of a slightly advanced kind, on the net? If so, on which sites can they be found?

Vinod Rathod said: (Sep 16, 2013)  
Lets do by this way :

As you said that (7c5*3c2) = (7c2*3c1).

7/5*3/2 = 7/2*3/1 = 63.

Pooja said: (Sep 28, 2013)  
I understood that thanks! we have to use nCr=nCn-r.

Chandan said: (Oct 30, 2013)  
Why multiplication not addition?

Aswin said: (Apr 2, 2014)  
Another question : how to select 6 people from a group of 10 so that two of them does not come together?

Satish said: (May 10, 2014)  
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)].

Shital said: (Dec 9, 2014)  
I didn't understand 7c2*3c1.

Vishesh said: (Dec 9, 2014)  
Same here what is this c? I don't understand this.

Manju said: (May 24, 2015)  
nCr = n!/(n-r)!r!
= 7C5*3C2.
= (7*6/2*1)*3 = 63.

Bond81 said: (Jun 23, 2015)  
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.

Bgt said: (Aug 8, 2015)  
How can't we use any other e example?

Sravani said: (Sep 24, 2015)  
How 3c2 = 3c1?

Harish Julai said: (Oct 7, 2015)  
Hello @Sravani.

If n>r, then ncr becomes nc(n-r).

Kunal said: (Nov 3, 2015)  
But why they did 7c5*3c2 as 7c2*3c1?

Why they directly not calculate 7c5*3c1?

Shivam Shukla said: (Dec 1, 2015)  
I didn't understand this problem. Please each and every step explain it.

Anoushka said: (Feb 26, 2016)  
Sir I haven't understood why have we multiplied 3 with 7*6/2*1 *3.

Sam said: (Jun 18, 2016)  
Why we are taking as 7c2 * 3c1? Please explain it.

Sudha said: (Sep 7, 2016)  
How 7c2 will come instead 7C5?

Mayank said: (Sep 23, 2016)  
@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.

Mariyam said: (Oct 13, 2016)  
@Satish your explanation is very understandable.

Dhayithri said: (Nov 11, 2016)  
Here we can also taken as 3c2 as 3c1 because factorial of those two is equal.

Feze said: (Dec 7, 2016)  
Why we multiply with 3 here?

Jyoti said: (Dec 15, 2016)  
According to me, the final answer which displayed is wrong.

If we multiply 126 is coming.

Shivaraj said: (Dec 18, 2016)  
@Feze.

3c1=3 By std formulae.

Naresh said: (Dec 20, 2016)  
How to get the answer? Please anybody explain me.

Thatchayini said: (Jan 11, 2017)  
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.

Kyosi Billy said: (Aug 26, 2017)  
Thanks for explaining it.

Bhavani said: (Mar 2, 2018)  
How will get 7c2* 3c1?

Please explain me.

Shree said: (May 21, 2018)  
Thanks for giving an explanation.

Mike said: (Jul 9, 2018)  
Where 6 comes from? Please explain me.

Ashutosh Sahoo said: (Aug 25, 2018)  
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).

Keerthana said: (Sep 11, 2018)  
The answer is 63. I agree.

Kamal said: (Sep 14, 2018)  
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?

Saideep Uikey said: (Oct 25, 2019)  
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.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.