Permutation/Combination....

A certain restaurent offers 6 kinds of cheese and 2 kinds of fruit for its dessert platter. If each dessert platter contains an equal number of kinds of cheese and kinds of fruit, how many different dessert platters could the restaurent offer?

A. 8
B. 12
C. 15
D. 21
E. 27

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givemeanid: Master | Next Rank: 500 Posts; Posts: 277; Joined: Sun Jun 17, 2007 2:51 pm; Location: New York, NY; Thanked: 6 times; Followed by:1 members

by givemeanid » Wed Jul 11, 2007 5:41 am

The restaurant can offer a platter with 1Cheese,1Fruit OR 2Cheese,2Fruits

1Cheese,1Fruit = 6C1*2C1
2Cheese,2Fruit = 6C2*2C2

Total number of platters restaurant can offer = 6C1*2C1 + 6C2*2C2 = 6*2 + 15*1 = 27

So It Goes

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jrbrown2: Master | Next Rank: 500 Posts; Posts: 103; Joined: Sun Jun 10, 2007 7:10 pm; Location: Brooklyn, NY; Thanked: 2 times

by jrbrown2 » Thu Jul 12, 2007 3:46 pm

The restaurant can offer a platter with 1Cheese,1Fruit OR 2Cheese,2Fruits

1Cheese,1Fruit = 6C1*2C1
2Cheese,2Fruit = 6C2*2C2

Total number of platters restaurant can offer = 6C1*2C1 + 6C2*2C2 = 6*2 + 15*1 = 27

Givemeanid,

Can you explain how to compute the 6C1, 6C2. I completely forgot how to compute those. And when is it necessary to use them?

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givemeanid: Master | Next Rank: 500 Posts; Posts: 277; Joined: Sun Jun 17, 2007 2:51 pm; Location: New York, NY; Thanked: 6 times; Followed by:1 members

by givemeanid » Thu Jul 12, 2007 5:45 pm

nCk = n! / (k! * (n-k)!)
nPk = n! / (n-k)!

So It Goes

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