Combinations?

Question

I've tried working this one out... and after two whole days have resorted to asking ANYONE to please please help me!


A committee of 3 is to be selected from 4 women and 5 men. The rules state that there must be at least one man and one woman on the committee.


a. In how many ways can the committee now be chosen?


b. Subsequently one of the men and one of the women marry each other. The rules also state that a married couple may not both serve on the committee. In how many ways can the committee now be chosen. (answer in the back of the book is 63)


Erm... could you also explain why you've done what you've done in the working because I just don't get combinations (the teacher wasn't in this week and set it as homework)

Answer 1

one man, two women = 5C1 times 4C2 = 5 X 6 = 30
two men, one woman = 5C2 times 4C1 = 10 X 4 = 40

So 70 total committees will have at least one of each gender

Of the 30 one man two women committees, the number having his wife in the committee is 1C1 times 3C1 which is 3

Of the 40 one woman two men committees, the number having her husband on them is 1C1 times 4 C 1 which is 4

Take these 7 committees away from the original 70 leaving 63.

Hope that helps.

Answer 2

a. First, we choose 1 man and 1 woman. There are 5*4 = 20 different pairs. Then we add another person; there are 9-2 = 7 choices. So it *seems* that there are 20*7 = 140 triples, but actually there has been double counting: each triple has been counted exactly twice. (Say the triple has two men, then each may have been chosen either "as a man" or as "the third person". In other words, if the men are called M and N, and the woman is W, then one and the same triple has appeared as MWN and also as NWM.) So actually the number of committees is 140/2 = 70.

b. It easier to count the complementary cases, that is, the number of triples with both of the married couple: this is just seven (there are 7 choices for the 3rd person). So the number of committees not including both of the couple is 70-7 = 63.

Answer 3

There are 6 combinations of 2 women times 5 men, therefore 6 X 5= 30
There are 10 combinations of 2 men times 4 women, therefore 10 X 4= 40

total combinations available 30 + 40= 70

In 3 instances a single women is involved in a combination of the women
In 4 instances a single man is involved in a combination of the men

Marry the two singles and remove the instances of occurances there fore 3 + 4= 7

total committies available without the married couple being alligned 70 - 7= 63

Answer 4

well for the 1st part lets consoder 3 cases
a committee of 3 have to be formed and there has to be 1 man and 1 women
so a committee can be like
a. 1 man and 2 women
b. 2 man and 1 women
no other cases fit the condition

now in 1st case no of ways of forming a committee are 5c1.4c2=30
in 2nd case no of ways are 5c2.4c1=40

so the answer is 30+40=70

abt the 2nd part i didnt get ur question can u plz elaborate it better

if the answer is wrong or u hv any doubt plz feel free to ask

Answer 5

a. _ _ _ = three blank spots on committee

Two ways of doing it are MMW and WWM only.
MMW
5 men and 2 spots means 5*2 + 4*1
4 women and 1 spot means 4*1
Add it all up = 18
WWM
4 women and 2 spots means 4*2 + 3*1
5 men and 1 spot means 5*1
Add it all up = 16

Add the two up = 34 ways to do it
I don't know how they came up with 63 for the second one. It should be less than the first.

I have to leave for class, so I don't have time to answer b., sorry.

Answer 6

a. C(4,1)*C(5,2)+C(4,2)*C(5,1)
where C is for combination
b.
case1-if married man is included
u have 3women &4 men to choose from
so u can do it by taking 1man & 1woman
or 2 women
then no. of ways of choosing
=C(3,1)*C(4,1)+C(3,2)
=15
case2-if married woman is included
u have 3women &4 men to choose from
so u can do it by taking 1man & 1woman
or 2 men
then no. of ways of choosing
=C(3,1)*C(4,1)+C(4,2)
=18
case3-if married couple is not included
u have 3women &4 men to choose from
so u can do it by taking 2man & 1woman
or 2 women & 1 man
then no. of ways of choosing
=C(3,1)*C(4,2)+C(3,2)*C(4,1)
=30
total no of ways=30+18+15
=63