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Counting books [#permalink] New post 10 Aug 2003, 08:55
Can someone please explain the solution to this problem.

There are 10 books on a shelf. 4 of them have a soft cover and 6 of them have a hard cover. In how many ways can you remove 5 books from the shelf so as to have at least one soft covered book and at least one hard covered book.
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 [#permalink] New post 10 Aug 2003, 10:24
5(all)-5(soft)-5(hard)
10C5-0-6C5

Correct? After Akamaibrah crucified me on prime factorization, I start to doubt in any problem.
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 [#permalink] New post 10 Aug 2003, 10:40
stolyar wrote:
5(all)-5(soft)-5(hard)
10C5-0-6C5

Correct? After Akamaibrah crucified me on prime factorization, I start to doubt in any problem.


Didn't mean to crush you on that problem. Just wanted to show your stubborn butt that giving someone a formula to get the answer is NOT the same as showing them how to solve the problem.

8-)


BTW, i got the same answer as you: 246. There is no reason to doubt when you have a rationale for coming up with your answer! :)
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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 [#permalink] New post 21 Jan 2004, 19:45
S H
0 5 = 6C5 = 6
1 4 = 4C1 * 6C4 = 60
2 3 = 4C2 * 6C3 = 120
3 2 = 4C3 * 6C2 = 60

Total = 246
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 [#permalink] New post 22 Jan 2004, 10:19
anandnk wrote:
S H
0 5 = 6C5 = 6
1 4 = 4C1 * 6C4 = 60
2 3 = 4C2 * 6C3 = 120
3 2 = 4C3 * 6C2 = 60

Total = 246


Anand the collection should have atleast one S and atleast one H.
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 [#permalink] New post 22 Jan 2004, 15:24
sorry about the mistake

S H

1 4 = 4C1 * 6C4 = 60
2 3 = 4C2 * 6C3 = 120
3 2 = 4C3 * 6C2 = 60
4 1 = 4C4 * 6C1 = 6

Total = 246
  [#permalink] 22 Jan 2004, 15:24
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