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?

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