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A promoter is organising an opera concert. Ten singers are

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Director
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A promoter is organising an opera concert. Ten singers are [#permalink] New post 08 Apr 2007, 23:57
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A promoter is organising an opera concert. Ten singers are available for the program: 3a's, 3m's, 2 t's and 2b's. If the promoter intends every segment to be a duet by performers with vocal ranges, how many pairings of performers are possible?

A 6
B 10
C 24
D 36
E 37
Director
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 [#permalink] New post 09 Apr 2007, 01:00
Hi ,
If i get your question correctly then we have ;
A&M-9 duets
A&T-6 duets
A&B-6 duets
M&T-6 duets
M&B-6 duets
T&B-4 duets

or 37 different duets in total
Senior Manager
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 [#permalink] New post 09 Apr 2007, 17:25
E?

3A*3M
3A*2T
3A*2B
total 21

3M*2T
3M*2B
total 12

2T*2B
total 4

total = 21+12+4 = 37
Manager
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Re: PS: Combinatorics [#permalink] New post 09 Apr 2007, 18:36
botirvoy wrote:
A promoter is organising an opera concert. Ten singers are available for the program: 3a's, 3m's, 2 t's and 2b's. If the promoter intends every segment to be a duet by performers with vocal ranges, how many pairings of performers are possible?

A 6
B 10
C 24
D 36
E 37


There are 4 groups. Number of ways to chose groups = 4C2 = 6

So we have am, at, ab, mt, mb and tb as possible combinations
Now we can chose 3x3 = 9 for am, 3x2 for at and so on. You will get 9 + 6 + 6 + 6 + 6 + 4 = 37 ways
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 [#permalink] New post 09 Apr 2007, 21:18
There are 6 groups we can get from 4 such groups --> 4C2 = 6

They are: am,at,ab,mt,mb,bt

For the first one, a&m, we have 3*3 = 9 combis
For the second one, a&t, we have 3*2 = 6 combis
For the third one, a&b, we have 6 combis
For the fourth one, m&t, we have 6 combis
For the fifth one, m&b, we have 6 combis
For the last one, b&t, we have 4 combis.

Total = 36 pairings.
Director
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 [#permalink] New post 10 Apr 2007, 02:37
Let me think this way:

Total number of singers = 10
The number of possible pairings = 10C2 = 45
Number of favorable pairings = Total pairings - Pairings of singers within the same group

Pairings of singers within the same group = 3C2 (from group 'a') + 3C2 (from 'm') + 2C2 ( from 't') + 2C2 (from 'b') = 8

Number of favorable pairings = 45-8 = 37
  [#permalink] 10 Apr 2007, 02:37
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