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?

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- March 27th 2008, 08:17 PMVulcanCommittees 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?

- March 27th 2008, 10:37 PMSoroban
Hello, Vulcan!

Quote:

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: . 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: ways.

We choose 2 more men from the remaining 5 men: ways.

Hence, there are: . committees with both*He*and*She*serving.

Therefore, there are: . committees

. . in which both*He*and*She*are not included.

- March 28th 2008, 10:58 AMVulcanCommittees and Combinations
Thanks Soroban.

- March 28th 2008, 11:01 AMVulcanJust 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?

- March 28th 2008, 11:13 AMiknowone
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? - March 28th 2008, 02:19 PMSoroban
Hello again, Vulcan!

Quote:

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?

*iknowone*had the best approach.

I intended to show the "long" way, but found that it is just as efficient.

His way . . .

There are: . possible pairs of books.

. . There are: . pairs of Math books.

. . There are: . pairs of Science books.

. . There are: . pairs of Economics books.

Hence, there are: . pairs of matching books.

Therefore, there are: . pairs of non-matched books.

The "long" way . . .

To choose 1 Math and 1 Science book: . ways.

To choose 1 Math and 1 Economics book: . ways.

To choose 1 Science and 1 Economic book: . ways.

Therefore, there are: . ways.

- March 29th 2008, 09:53 AMVulcan
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?