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Combination problem

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Combination problem [#permalink] New post 18 Jul 2006, 06:28
Can anyone please answer and explain the following question -

Thanks

--
To furnish a room in a model house, an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tables in a warehouse that are all different from each other. If there are 5 chairs in the warehouse and if 150 different combinations are possible, how mnay tables are in the warehouse?
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 [#permalink] New post 18 Jul 2006, 07:14
150 = 5C2 * nC2

Solving this we get n=6 hence 6 tables.
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 [#permalink] New post 18 Jul 2006, 07:52
I did it by trial n error and got 6 tables.

6C2.
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 [#permalink] New post 18 Jul 2006, 18:10
total possible combination = 150

from selecting chiars 5C2
from selecting tables XC2

thus 5C2*XC2 = 150

(5*4)/2 * (X*X-1)/2 = 150

X^2-X-30 = 0

X = 6 or -5 (X>0)

so ans is 6
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 [#permalink] New post 18 Jul 2006, 18:17
# of ways to pick 2 chairs = 5C2 = 5!/2!3! = 10 ways

# of ways to pick tables = tC2 = t!/2!(t-2)!

Total number of ways = 150 = 10 * t!/(t-2)!2!

15 = t!/2!(t-2)!
30 = t!/(t-2)!
t(t-1) = 30
t^2 -t - 30 = 0

t = 6
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 [#permalink] New post 18 Jul 2006, 22:01
Two chairs can be chosen in 5C2 ways
Let there be n tables.
Two tables can be chosen in nC2 ways.

Total ways = 5C2*nC2

Given there are 150 ways

5C2*nC2 = 150
10 * nC2 = 150
nC2 = 15

hence n = 6
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 [#permalink] New post 18 Jul 2006, 22:12
5C2 * xC2 = 150

10 * x * (x-1)/2 = 150
x*(x-1) = 30
x^2-x-30= 0
(x-6)(x-5) = 0
Therefore x = 6
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  [#permalink] 18 Jul 2006, 22:12
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