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.

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).

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.

The answer is 63. I agree.

Thanks for giving an explanation.

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)].

Thank you so much @Satish.

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?

How to get the answer? Please anybody explain me.

How 7c2 will come instead 7C5?