Veritas Probability

What's the slot method??

Any other method besides the one that I've used.lol

Ok. May be you are looking for this process then..

Two cases :
1. when either of the two members is included.
=5C3*2C1 (Out of 2 members ways of selecting one of the two members=2C1 and out of the remaining 5 members ways of selecting 3 members)
=20 cases
2. When none of the two members are included
= 5C4 (no of ways of selecting 4 members out of 5 people)
=5

Therefore total number of ways of selecting 4 people out of 7 when 2 wont work together= 20+5=25

knight247 wrote:A committee of 4 is to be chosen out of 7 employees for a special project at ACME corporation. 2 of the 7 employees are unwilling to work with each other. How many committees are possible if the 2 employees do not work together?

Fairly simple one. Find the total of possibilities w.o restriction and from that subtract the possibilities where the 2 employees will always be together i.e.7C4-5C2=35-10=25

However, for the sake of getting a thorough understanding, was hoping that someone could show me how to do this using a different method i.e. the straight forward method like the slot method. I know its more time consuming but would like to sharpen my math skills. Thanks

Ok I got it.

1. Number of possibilities where neither of those two enemy employees will be there
That can happen in 5C4 ways=5 ways

OR

2. Number of possibilities where there will be only one of those two enemy employees and the rest will be from among the other 5 normal employees=2C1*5C3=2*5=20 ways

So we have a total of 5+20=25ways.