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elmiko
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 Posted: Sat Oct 11, 2008 3:25 pm Post subject: Tricky Combination Problem |
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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?
Answer: 150
Thanks in advance! |
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ab78
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| Posted: Sat Oct 11, 2008 4:21 pm Post subject: |
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| Is the answer 90? |
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nitin86
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| Posted: Sat Oct 11, 2008 7:32 pm Post subject: |
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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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cramya
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| Posted: Sat Oct 11, 2008 7:39 pm Post subject: |
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| Good explanation Nitin! |
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4meonly
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| Posted: Sun Oct 12, 2008 7:48 am Post subject: |
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| 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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ab78
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| Posted: Sun Oct 12, 2008 8:08 am Post subject: |
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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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parallel_chase
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| Posted: Sun Oct 12, 2008 9:42 am Post subject: |
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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.
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