# Probability question: From a group of 8 women, 6 men,

• Jamin2112
In summary: Then you can see how many of them contain the forbidden combination.In summary, the number of possible committees that can be formed from a group of 8 women and 6 men, with 3 men and 3 women on each committee, is (8 choose 3) * (6 choose 3). To find the number of committees where two specific men cannot serve together, we can use the elimination method and subtract the total number of illegal committees from the total number of possible committees. This can also be done by enumerating all possible legal combinations of three men and counting how many of them contain the forbidden combination. The final result is 96 possible committees.
Jamin2112
Probability question: "From a group of 8 women, 6 men, ..."

## Homework Statement

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 [...] 2 of the men refuse to serve together?

## Homework Equations

n choose k = n!/((n-k)! * k!)

multiplication rule

## The Attempt at a Solution

Clearly the number of total committees that can be formed from 3 of 8 women and 3 of 6 men is (8 choose 3) * (6 choose 3). I need to subtract the number of those committees in which some arbitrary pair of men would've served together. Could someone please explain (or hint at) the logic involved in finding this?

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I would do this: call the 6 men a, b, c, d, e, and f and assume, without loss of generality, that the two men who refuse to serve on the same committee are a and b. Then there are three ways to choose the three men to serve on the commitee:
1) Choose a but not b. Then we must choose 2 more men from the four, c, d, e, and f. There are 4(3)= 12 ways to do that.
2) Choose b but not a. As before, there are 12 ways to do that.
3) Choose neither a nor b to be on the committee. That is, we choose 3 out of c, d, e, and f. There are 4(3)(2)= 24 ways to do that.

There are, therefore, 12+ 12+ 48= 96 ways to choose the three men.

There are, as you say. "8 choose 3"= 8(7)(6)= 336 ways to choose the three women.

1) Choose a but not b. Then we must choose 2 more men from the four, c, d, e, and f. There are 4(3)= 12 ways to do that.
2) Choose b but not a. As before, there are 12 ways to do that.
3) Choose neither a nor b to be on the committee. That is, we choose 3 out of c, d, e, and f. There are 4(3)(2)= 24 ways to do that.

There are, therefore, 12+ 12+ 48= 96 ways to choose the three men.
I disagree here; order doesn't matter for any group of men, and as such the choose function should show up here.

As far as the OP's elimination method goes, again, order does not matter. How many 'illegal' committees are there? (Hint: think about how many more men there are to choose from, and how big a committee is.)

Alternatively, using the letter system, it's actually not unrealistic to enumerate all possible 'legal' combinations of three men (though I'll grant it's rather tedious).

## What is the probability of randomly selecting 3 women from the group?

The probability of randomly selecting 3 women from a group of 8 women and 6 men is 8/14, or approximately 57.14%.

## What is the probability of randomly selecting 2 men and 1 woman from the group?

The probability of randomly selecting 2 men and 1 woman from a group of 8 women and 6 men is (6/14) x (5/13) x (8/12), or approximately 18.46%.

## What is the probability of selecting at least 1 man from the group?

The probability of selecting at least 1 man from a group of 8 women and 6 men is equal to 1 - the probability of selecting only women. This is equal to 1 - (8/14), or approximately 42.86%.

## What is the probability of randomly selecting 3 people from the group and having at least 2 women?

The probability of randomly selecting 3 people from a group of 8 women and 6 men and having at least 2 women is equal to the sum of the probabilities of selecting 2 women and 3 women. This is (8/14) x (7/13) x (6/12) + (8/14) x (7/13) x (6/12) x (5/11), or approximately 52.38%.

## What is the probability of selecting all women or all men from the group?

The probability of selecting all women or all men from a group of 8 women and 6 men is equal to the sum of the probabilities of selecting all women and selecting all men. This is (8/14) x (7/13) x (6/12) + (6/14) x (5/13) x (4/12), or approximately 5.95%.

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