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# In how many ways can 5 identical black balls and 7 identical

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Director
Joined: 19 Nov 2004
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In how many ways can 5 identical black balls and 7 identical [#permalink]

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06 Jan 2005, 10:47
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In how many ways can 5 identical black balls and 7 identical white balls be arranged in a row so that no 2 black balls are together

A)56
B)64
C)65
D) 316
E)560

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Director
Joined: 19 Nov 2004
Posts: 556

Kudos [?]: 272 [0], given: 0

Location: SF Bay Area, USA

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07 Jan 2005, 12:24
Comon guys ...
I am bumping this up the list, to get more attention

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Director
Joined: 07 Nov 2004
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08 Jan 2005, 09:12
Chose A.

The only way you can place the 5 black balls without have two together would be intersperse them between the white balls.

_W_W_W_W_W_W_W_

If you look at the above arrangement there are 8 possible locations for the 5 black ball to go and since they are identical, the order in the which they appear is not important. So the answer is 8C5 = 56

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Manager
Joined: 17 Dec 2004
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Location: Find me if you can

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08 Jan 2005, 11:24
I agree, this is the best approach.

56
_________________

Wish you all good luck
Bhimsen Joshi

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Manager
Joined: 31 Aug 2004
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10 Jan 2005, 00:09
gayathri wrote:
Chose A.

The only way you can place the 5 black balls without have two together would be intersperse them between the white balls.

_W_W_W_W_W_W_W_

If you look at the above arrangement there are 8 possible locations for the 5 black ball to go and since they are identical, the order in the which they appear is not important. So the answer is 8C5 = 56

gayathri, I still have difficulty to understand your approach.

There are 7 spaces between the 7 white balls plus the one ahead of the first white one, so there are 8 spaces available for the 5 black balls to fit.

How does this correspond to the logic of 8C5, 8 choose 5?

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Director
Joined: 07 Nov 2004
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10 Jan 2005, 07:14
jinino wrote:
gayathri wrote:
Chose A.

The only way you can place the 5 black balls without have two together would be intersperse them between the white balls.

_W_W_W_W_W_W_W_

If you look at the above arrangement there are 8 possible locations for the 5 black ball to go and since they are identical, the order in the which they appear is not important. So the answer is 8C5 = 56

gayathri, I still have difficulty to understand your approach.

There are 7 spaces between the 7 white balls plus the one ahead of the first white one, so there are 8 spaces available for the 5 black balls to fit.

How does this correspond to the logic of 8C5, 8 choose 5?

You have 8 empty positions but you have only 5 black balls to fill the space. According the question only the black balls should not be next to each other there is no such restriction for the white balls. So, now the question is similar to how many sets of 5 can you choose from 8?

You use combinations instead of permutation because order does not matter as all the black balls are identical. If the balls had been numbered then you would have done 8P5.

HTH

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10 Jan 2005, 07:14
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