Aptitude - Permutation and Combination - Discussion

I understood all the combination how we choose, Thanks for this @ Mohan & @Pavz. :)

But I have little doubt in solving cases like;

1) why we "multiple" in case?

2) why we "add" all cases?

The question is not specific regarding the selection of men and women after the selection of at least three men.

If we take the method [7C3 * 6C2] + [7C4 * 6C1] + [7C5 * 6C0] as the solution, then we must understand that all men are taken as one entity first then women are considered as another entity.

Hence, 3 men are one entity and 2 women are another entity. Similarly, 4 men are one and likewise, 5 men are one entity.

That means if you are done with the selection of 3 men at least, then DONT SELECT MEN AGAIN from the remaining. And if you choose to select men again, THEN INCLUDE THEM WHILE SELECTING ALTEAST THREE MEN.

Why can't we select only five women out of 6 women. Why that combination is not taken into account.

I am still have little problem with the solving.

I am still have little problem with the solving.

In that question that is cleared that 5 members need to be select out of 7 men and 6 women.

And in question there is a condition that at least 3 men that mean we can select in 3 ways that we can select 3 men or 4 men or 5 men.

1) 3 men and we want another 5-3 = 2 women. We have to select 3 men from 7 men that is 7C3 and 2 womens from 6 women that is 6C2 results is 7C3*6C2.

2) 4men and we want another 5-4 = 1 women. We have to select 4 men from 7 men that is 7C4 and 1 women from 6 women that is 6C1 results is 7C4*6C1.

3) 5 men and we want another 5-5 = 0 women. We have to select 5 men from 7 men that is 7C5 and 0 women from 6 women that is 6C2 results is 7C5*6C0.

Answer is we need to sum up all these combinations that is (7C3*6C2+7C4*6C1+7C5*6C0) that is 756.

How 6C2 equal to 15 please explain?

Why 7C3*10C2 is wrong?

Lets take a case:

We have 7 men namely A B C D E F G and 6 women H I J K L M.

Case 1) 3 men and 2 from the set of 10 = A B C + D H.

Case 2) 3 men and 2 from the set of 10 = A B D + C H.

Both the sets are identical. This shows that our sets are not mutually exclusive. Hence we must choose men and women separately.

I used combination formula, nCr = n!/r! (n-r)!

Required number of ways = (7C3*6C2)+(7C4*6C1)+7C5.

= (7*6*5/3*2*1+6*5*1/2*1)+(7*6*5*4 /4*3*2*1+6/1)+(7*6*5*4*3*2*1/5*4*3*2*1).

= 35*15+35*6+24 = 756 answer.

How many 3 digit numbers can be formed from 2, 3, 4, 5, 6, 7, 9, which are divisible by 5 and none of the digits is repeated? Please tell answer and method for this problem.