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Tricky Combination Problem

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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!
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by ab78 » Sat Oct 11, 2008 4:21 pm
Is the answer 90?
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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) ]
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by cramya » Sat Oct 11, 2008 7:39 pm
Good explanation Nitin!
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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.
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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.
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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....
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