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combination

by Guest Thu Mar 27, 2008 7:26 am

7)To furnish a room in model home,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 many tables are in the warehouse?

a 6 b 8 c 10 d 15 e 30
oa is a.6

plz help how to solve this.
jc
 
 

Re: combination

by jc Thu Mar 27, 2008 8:40 pm

Anonymous Wrote:7)To furnish a room in model home,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 many tables are in the warehouse?

a 6 b 8 c 10 d 15 e 30
oa is a.6

plz help how to solve this.



Formula = nCr = n!/r!(n-r)! , which will give u the number of ways of choosing r items out of n items.

No of ways of choosing 2 chairs out of 5 and 2 tables out of x tables = 150

ie, 5C2 . xC2 = 150

5!/2!.3! . x!/2!.(x-2)! = 150

(x.(x-1).(x-2)!)/(x-2)! = 30

x^2 - x -30 =0

Solving , we get x=6 or x=-5
RonPurewal
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Joined: Tue Aug 14, 2007 8:23 am
 

Re: combination

by RonPurewal Sat Mar 29, 2008 12:54 am

jc Wrote:
Anonymous Wrote:7)To furnish a room in model home,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 many tables are in the warehouse?

a 6 b 8 c 10 d 15 e 30
oa is a.6

plz help how to solve this.



Formula = nCr = n!/r!(n-r)! , which will give u the number of ways of choosing r items out of n items.

No of ways of choosing 2 chairs out of 5 and 2 tables out of x tables = 150

ie, 5C2 . xC2 = 150

5!/2!.3! . x!/2!.(x-2)! = 150

(x.(x-1).(x-2)!)/(x-2)! = 30

x^2 - x -30 =0

Solving , we get x=6 or x=-5


this is a good solution.

note that once you get to xC2 = 15, it suffices to just test different values for x until you get the magic 15, rather than to transform the equation into a quadratic.
given that you just found out 5C2 = 10, it's clear that the 'x' you're looking for is just a lil greater than 5. 6, therefore, is the natural first choice to check, and it works.
Suyash
 
 

by Suyash Sat Mar 29, 2008 1:40 pm

Thank you jc and ron.