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Combinatorics

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by kishore » Sat Jun 21, 2008 6:58 pm
Easy way is to go by answers


5C2 * XC2 = 150

If X = 6 , it solves the above equation


so, answer is 6
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by rosh26 » Sat Jun 21, 2008 8:24 pm
Is 5C2 * XC2 = 150 = 5!/2! * X!/2! ??

Please confirm.. Thanks.
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by kishore » Sat Jun 21, 2008 8:51 pm
No.....it is

nCr = n!/((n-r)! * r!)

5C2 = 5!/(3! * 2!)

XC2 = X!/((X-2)! * 2!)

Substitute X = 6

6C2 = 6!/((6-2)! * 2!)
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by atlantic » Sun Jun 22, 2008 10:38 am
Rosh,

nr. of ways arrange chairs * nr. of ways arrange tables = 150

nr. of ways arrange chairs= 5!/(2!(5-2)!)=5!/(2!3!)=10

So, nr. of ways arrange tables = 15

If x is the number of tables available at the store,

x!/(2!(x-2)!) = 15

It's now time to test given solutions.

6!/(2!4!)=15, so A it is.
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