Permutation and combination

Question

Answer 1

We have a total of 15 members in the committee, consisting of 7 men and 8 women.

Now both men and women needs to be assigned for a total of 3 tasks.

These tasks are assigned in such a way that at least both a man and a woman must be assigned a task.

Therefore the cases that can be formed are:

Case 1: 

Task 1 : 1 Man

Task 2 : 1 Man

Task 3 : 1 Woman

Case 2: 

Task 1 : 1 Man

Task 2 : 1 Woman

Task 3 : 1 Woman

[Please note there will be no cases such that we assigned 1 Man to task 1, 1 Woman to task 2 and leave the task 3 as at least a man and a woman are assigned. Every task needs a person.]

All now left is to calculate the separate cases and sum up them.

Case 1: To choose 2 men out of 7 men and 1 woman out of 8 women. 

(7 choose 2) * (8 choose 1) = 21 * 8 = 168

Case 2: To choose 1 men out of 7 men and 2 woman out of 8 woman

(7 choose 1) * (8 choose 2) = 7 * 28 = 196

Therefore total ways = 168 + 196 = 364 ways.

Hence, there are 364 ways to complete the tasks.

Comments

we can also do it by bijection choosing three person out of 15 subtracting all cases when all 3 boys or 3 girls are choosen.......15C3 -7C3-8C3= 364

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i think task also be  different  so these are multiply with 3   i am right?

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I think the problem can be looked at like this:

First we pick two tasks out of the 3 in C(3,2) ways. Then we give one of those tasks to men and the other task to women. It can be done in 7*8 = 56 ways. Then we consider the possibility where the jobs are assigned in the reverse order. That would be another 56 ways. Then we have one task remaining which can be assigned to any one. Therefore 15 ways. Hence the answer can be

C(3,2) * 2 * 56 * 15 = 5040 ways.

Please correct me if I am wrong.