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# In a certain company, three typists service for four

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Senior Manager
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In a certain company, three typists service for four [#permalink]

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09 Mar 2008, 10:21
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In a certain company, three typists service for four departments. If the departments each send a file to the typists at random, what is the probability that every typist will recieve at least one file?

A. 8/9
B. 64/81
C. 4/9
D. 16/81
E. 5/9

Spoiler: C
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Senior Manager
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13 Mar 2008, 18:48
This question got burried in the pile...good one though

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13 Mar 2008, 23:01
(4*4C3)/81=16/81
D
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14 Mar 2008, 06:38
oana2000 wrote:
(4*4C3)/81=16/81
D

Can you explain 4*4C3?
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14 Mar 2008, 09:50
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This is how I have done it:

Let's say x, y and z are the typists - so if all of them have to get atleast one file, one of them gets two. So the pattern is 1,1,2 or 1,2,1 or 2,1,1

Let us take 1,1,2 - files can be sent to z in 4c2 ways, and teh remaining files can be sent to x and y in 2 ways. So that makes the total combinations = 2*4c2 = 12

If we add numbers for 1,2,1 and 2,1,1 - total becomes 36

So probability = 36/81 = 4/9
Re: typists probability   [#permalink] 14 Mar 2008, 09:50
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# In a certain company, three typists service for four

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