Aptitude - Permutation and Combination - Discussion

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.

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)].

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.

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.

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.

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.

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?

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?

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.

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).