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In how many ways can one post 5 letters in 10 different

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In how many ways can one post 5 letters in 10 different [#permalink] New post 25 Jan 2004, 17:08
In how many ways can one post 5 letters in 10 different letter boxes?

100000
9765625
15
50
27
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 [#permalink] New post 26 Jan 2004, 05:17
Any of the letters we can post in 10 ways so 5 letters can be posted in 10^5
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 [#permalink] New post 26 Jan 2004, 14:59
:cool
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 [#permalink] New post 29 Jan 2004, 10:51
sunniboy007 wrote:
:cool


Just an extension of the problem. In how many ways can one post 5 letters in 10 different letter boxes if no more than one letter should be posted in the same letter box?
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 [#permalink] New post 29 Jan 2004, 11:00
sunniboy007 wrote:
:cool


Hey Sunnyboy,


I think the answer should be 5^10 which is second choice. Are you sure that the official answer is 100000?

Thanks
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 [#permalink] New post 29 Jan 2004, 13:36
Ofcourse sono sicuro che ├и la risposta :lol:
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 [#permalink] New post 29 Jan 2004, 13:38
sunniboy007 wrote:
Ofcourse sono sicuro che ├и la risposta :lol:


I do not understand what you wrote. What is the answer?
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 [#permalink] New post 29 Jan 2004, 13:40
I said, Of course I am sure that 100000 is the correct answer.
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 [#permalink] New post 30 Jan 2004, 19:36
sunniboy007 wrote:
I said, Of course I am sure that 100000 is the correct answer.



Ok try to draw similarity with this problem. How many different values can 3 bits represent if each bit has two possible values?
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 [#permalink] New post 30 Jan 2004, 19:54
Once you post a letter in a box, you cant use the letter posted in the calculation.

This problem sounds more or like #partitions of a number.

Find #non-negative integer solutions of the following eqn.

k1 + k2 + .... + k10 = 5

Answer is 2002
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 [#permalink] New post 04 Feb 2004, 02:12
gmatblast wrote:
sunniboy007 wrote:
:cool


Just an extension of the problem. In how many ways can one post 5 letters in 10 different letter boxes if no more than one letter should be posted in the same letter box?


think this way. You have 10 choices for the first letter, then 9 choices for the second letter, the 8 choices for the third letter.

Hence, you have 10*9*8*7*6 = 30240.

This can also be thought of as an "arrangement" or "permutation" problem.

How many ways can you assign a mailbox number to 5 different letters without repeating from a choice of 10 numbers. 10P5 = 10!/5! = same solution as above.
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MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

  [#permalink] 04 Feb 2004, 02:12
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