1. ## Combination problems

1. Five separate awards are to be presented to selected students of a class of 30. how many different outcomes possible if
a. a student ca receive any number of awards?
b. each student can receive at most 1 award?

2. A student has to sell 2 books from a collection of 6 math, 7 science, and 4 economics books How many choices are possible if the books are to be on different subjects?

3. From a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is tone formed. How many different committees are possible if 1 man and 1 woman refuse to serve together?

4. A person has 8 friends of whom 5 will be invited to a party. How many choices are there if 2 of the friends will only attend together?

2. ## Re: Combination problems

1.a. 5^5 = 31125
1.b. 5! = 60

2. There 6*7 - 42 different ways to sell 1 math and 1 science book, 6*4 = 24 ways to sell 1 math and 1 economics book, and 7*4 = 28 ways to sell 1 science and 1 economics book, for a total of 94.

3. Lets call the 2 people who will not serve together John and Mary. There are (8 choose 3) * (6 choose 3) = 1120 total committees, and (7 choose 2) * (5 choose 2) = 210 that that have both John and Mary, so there are 1120 - 210 = 910 possible committees that do not include both John and Mary. Or you could also calculate the number of committees that Mary is on but not John (7 choose 2) * (5 choose 3) = 210, plus the number of committees that John is on but not Mary (5 choose 2) * (7 choose 3) = 350, plus the number of committees that both Mary and John are not on (7 choose 3) * (5 choose 3) = 350, for a total of 910.

4. Lets call the 2 friends who will only attend together John and Mary. There are (6 choose 5) = 6 ways to invite 5 friends that would not include John or Mary, and there are (6 choose 3) = 20 ways to choose 5 friends that would include both John and Mary, so there are 26 choices for invites to the party. Or you could approach by calculating the total number of invites possible (8 choose 5) = 56 and subtracting the number that would include only John (6 choose 4) = 10 and the number that would include only Mary (6 choose 4) = 10. So the total would e 56 - 10 -10 = 26.

I am trying to get better at these types of problems myself, so above is what I think are correct answers but not certain myself.

3. ## Re: Combination problems

Apollo, after being on this forum for a few days now, I realize I should not have simply provided answers. The idea as I have learned is for you to take an honest shot at solving, show where you are running into trouble, and other posters can help you/suggest an approach that will help you solve the problems. I did at least try to explain my logic on most of the problems so hopefully this helps on future problems, but as I have learned on future posts you should not just post questions, show your attempt to solve and where you run into a problem.

Hope to see more posts from you in the future and hope I can be of assistance in the future