BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Veritas Probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pm
Thanked: 114 times
Followed by:11 members

Veritas Probability

by knight247 » Tue Sep 06, 2011 12:44 am
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
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1309
Joined: Mon Apr 04, 2011 5:34 am
Location: India
Thanked: 310 times
Followed by:123 members
GMAT Score:750

by cans » Tue Sep 06, 2011 2:09 am
What's the slot method??
If my post helped you- let me know by pushing the thanks button ;)

Contact me about long distance tutoring!
[email protected]

Cans!!
Top
User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pm
Thanked: 114 times
Followed by:11 members

by knight247 » Tue Sep 06, 2011 3:15 am
Any other method besides the one that I've used.lol
Top
Senior | Next Rank: 100 Posts
Posts: 47
Joined: Thu Jul 14, 2011 9:41 am
Thanked: 9 times
GMAT Score:650

by prashant.mishra » Tue Sep 06, 2011 3:44 am
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
Top
User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pm
Thanked: 114 times
Followed by:11 members

by knight247 » Tue Sep 06, 2011 3:50 am
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.
Top
Post new topic Post Reply
• Page 1 of 1