# Math Help - Exams revision, help needed

1. ## Exams revision, help needed

A commision of 5 person is to be chosen from 6 men and 4 women. In how many ways can this be done,

1 if there must be 3 men and 2 women

2 more men than women

3 there must be 3 men and 2 women and one particular woman refuses to be on the comittee with one particular man ?

I did this

6 x 5 x 4 x 4 x 3 = 1440, doesn't seem to be right answer, 120 is the answer, could anyone explain me this and when to use combination and permutation, Thank you very much

2. Originally Posted by ramzel
A commision of 5 person is to be chosen from 6 men and 4 women. In how many ways can this be done,

1 if there must be 3 men and 2 women

2 more men than women

3 there must be 3 men and 2 women and one particular woman refuses to be on the comittee with one particular man ?

I did this

6 x 5 x 4 x 4 x 3 = 1440, doesn't seem to be right answer, 120 is the answer, could anyone explain me this and when to use combination and permutation, Thank you very much
1. ${6 \choose 3} \times {4 \choose 2} = ....$.

2. (4M and 1W) + (3M and 2W). Do the same sort of thing as above for each case.

3. (Number of possible committees that have the unwanted man) + (number of possible committes that don't have the unwanted man).

For the first case you're choosing 3 men from the 6 and 2 women from 3 (not 4). For the second case you're choosing 3 men from 5 (not 6) and 2 women from 4. For each case do the same sort of thing as was done in question 1.

3. Thank you but for the second part i dont seem to understand why am i choosing from 5 ? why do you always like choose one less ?