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

Redeem

Tricky Combination Problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Tue Sep 16, 2008 3:22 pm

Tricky Combination Problem

by elmiko » Sat Oct 11, 2008 3:25 pm
Hi all,

Came across the following question today that's driving me crazy:

An engagement team consists of a project manager, team leader, and 4 consultants. There are 2 candidates for PM, 3 candidates for TL, and 7 candidates for the 4 consultant positions. If 2 of the 7 consultants refuse to be on the same team, how many different teams are possible?

[spoiler]Answer: 150[/spoiler]

Thanks in advance!
Top
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Mon Sep 29, 2008 12:24 pm
Thanked: 1 times

by ab78 » Sat Oct 11, 2008 4:21 pm
Is the answer 90?
Top
Master | Next Rank: 500 Posts
Posts: 111
Joined: Sat Sep 01, 2007 11:00 pm
Thanked: 4 times

by nitin86 » Sat Oct 11, 2008 7:32 pm
No. of ways to select a PM = 2C1
No. of ways to select a TL = 3C1

No. of ways to select a consultants =>

(a) Both the consultants who don't wants to be on the same team are not selected, then number of ways to select 4 from ( 7 - 2) is

5C4

(b) Only one of the two consultants who don't want to be on the same team is selected, then number of ways is

(2C1) * (5C3) => (No. of ways to select one of the 2 consultant ) * ( No of wayst to select remaining 3 consultant from 5 , removing the 2 consultant..)

Hence, total ways

=> (2C1) * (3C1) * [ 5C4 + (2C1) * (5C3) ]
Top
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pm
Thanked: 331 times
Followed by:11 members

by cramya » Sat Oct 11, 2008 7:39 pm
Good explanation Nitin!
Top
Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 am
Thanked: 27 times
Followed by:1 members
GMAT Score:660(

by 4meonly » Sun Oct 12, 2008 7:48 am
nitin86 wrote:No. of ways to select a PM = 2C1
No. of ways to select a TL = 3C1

No. of ways to select a consultants =>

(a) Both the consultants who don't wants to be on the same team are not selected, then number of ways to select 4 from ( 7 - 2) is

5C4

(b) Only one of the two consultants who don't want to be on the same team is selected, then number of ways is

(2C1) * (5C3) => (No. of ways to select one of the 2 consultant ) * ( No of wayst to select remaining 3 consultant from 5 , removing the 2 consultant..)

Hence, total ways

=> (2C1) * (3C1) * [ 5C4 + (2C1) * (5C3) ]
nitin86, perfect!

I think we also can find the answer by following logic
(total number of combinations) - (combinations when 2 consultans are in the same group)

2C1*3C1*7C4 - 2C1*3C1*5С2 =210 - 60 = 150
Last edited by 4meonly on Sun Oct 12, 2008 11:23 am, edited 1 time in total.
Top
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Mon Sep 29, 2008 12:24 pm
Thanked: 1 times

by ab78 » Sun Oct 12, 2008 8:08 am
aah here is where i went wrong : while picking consultants I picked up

5C4 + 5C3..should have done "and" with 2C1 in the second part.

that was a good explanation.
Top
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 am
Thanked: 146 times
Followed by:2 members

by parallel_chase » Sun Oct 12, 2008 9:42 am
Here is another way: (probably simpler)

total no. ways the PM, TL and Consultants can be selected

2C1 * 3C1 * 7C4 = 2*3*35 = 210

No. of ways both consultants are selected together in one team

2C1 * 3C1 * 5C2 = 2*3*10 = 60

5C2 [ 2 consultants are already selected for the same team, out of remaining 5 we just have to select another 2]

No. of ways both the consultants are not selected in one team = 210-60 = 150


Hope this helps.
No rest for the Wicked....
Top
Post new topic Post Reply
• Page 1 of 1