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by drizzle » Sun Aug 10, 2008 7:33 am
case 1: 2 different size + 4 different colors...so for each color there are two ways ..total = 8 ways

case 2: 4C3 * 2 = 8

total = 8 + 8= 16... wht's the answere?
guess i'm not missing anything...
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by Senator 153 » Sun Aug 10, 2008 7:06 pm
I agree with drizzle's answer and method. I scribbled a quick list of each basic color-size pair ("blue green yellow pink blue-D green-D yellow-D pink-D) to help myself think. You have to know the combinations formula for this question: 4C3 = 4!/(3!)(4-3)! = 4.

Btw Veritas has a nice nice book on permutations and combinations on their online store, I used it last month and found it really helpful.
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by anju » Thu Aug 14, 2008 11:47 am
drizzle wrote:case 1: 2 different size + 4 different colors...so for each color there are two ways ..total = 8 ways

case 2: 4C3 * 2 = 8

total = 8 + 8= 16... wht's the answere?
guess i'm not missing anything...
I got the answer by anagram. Can you pls elaborate on why you took 4C3.

Thanks
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