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by senthil » Wed Apr 02, 2008 9:04 am
I think the answer is 6 . If its right below is my approach .
Out of 5 chairs 2 chairs are to be chosen = 5C2 =10 combination possible
Out of N tables 2 tables are to be chosen= NC2 combination possible

Since all the tables and chairs are different , the Total number of combinations (i.e. 150 ) is equal to 5C2*NC2
((n(n-1)) /2 )*10 = 150
solving we get a quadratic equation and again solving we get N = 6.

If I am wrong let me know
Thanks
Senthil
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by Kaunteya1 » Wed Apr 02, 2008 9:55 am
Number of chair combos can be determined, there are 5 and you need 2.

5c2 = 10
There are a total of 150 combinations. To get 150 you have to multipy the number of chair combos x the number of table combos.

5c2 x (table combos) = 150
10 x (table combos) = 150
table combos = 15

Nc2=15
the only way for that to be possible is 6c2 or 6!/(2!)(4!) or (6x5)/2=15

There are 6 tables.

This way you don't have to deal with quadratic.
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