1. ## Committees and Combinations

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

2. Hello, Vulcan!

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

With no restrictions, there are: .$\displaystyle {8\choose3}{6\choose3} \:=\:56\cdot20 \:=\:1120$ possible committees.

We will count the cases in which both He and She are on the committee.

He and She are already on the committee.
We choose 2 more women from the remaining 7 women: $\displaystyle {7\choose2} \:=\:21$ ways.
We choose 2 more men from the remaining 5 men: $\displaystyle {5\choose2} \:=\:10$ ways.

Hence, there are: .$\displaystyle 21 \times 10 \:=\:210$ committees with both He and She serving.

Therefore, there are: .$\displaystyle 1120 - 210 \:=\:{\color{blue}910}$ committees
. . in which both He and She are not included.

3. ## Committees and Combinations

Thanks Soroban.

4. ## Just one more

I've got one more problem like this. 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?

5. How many ways are there to sell any 2 books, total?

How many ways are there to sell 2 books if we require they be the same type?

So how many ways are there to sell 2 books if we require they be different?

6. Hello again, Vulcan!

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?
I assumed that iknowone had the best approach.
I intended to show the "long" way, but found that it is just as efficient.

His way . . .
There are: .$\displaystyle {17\choose2} \:=\:136$ possible pairs of books.
. . There are: .$\displaystyle {6\choose2} \:=\:15$ pairs of Math books.
. . There are: .$\displaystyle {7\choose2} \:=\:21$ pairs of Science books.
. . There are: .$\displaystyle {4\choose2} \:=\:6$ pairs of Economics books.
Hence, there are: .$\displaystyle 15 +21 + 6 \:=\:42$ pairs of matching books.

Therefore, there are: .$\displaystyle 136 - 42 \;=\;\boxed{94}$ pairs of non-matched books.

The "long" way . . .
To choose 1 Math and 1 Science book: .$\displaystyle {6\choose1}{7\choose1} \:=\:6\cdot7\:=\:42$ ways.
To choose 1 Math and 1 Economics book: .$\displaystyle {6\choose}{4\choose1} \:=\:6\cdot4 \:=\:24$ ways.
To choose 1 Science and 1 Economic book: .$\displaystyle {7\choose1}{4\choose1} \:=\:7\cdot4\:=\:28$ ways.

Therefore, there are: .$\displaystyle 42 + 24 + 28 \:=\:\boxed{94}$ ways.

7. Hmmmmm...
I understand the math iknowone used, but I don't understand why he used it. If the problem is about the number of groups-of-one you could create, why bother with groups-of-two?