- #1

Jamin2112

- 986

- 12

**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?

*OT link removed*

Last edited by a moderator: