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

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In how many ways can 5 identical black balls and 7 identical [#permalink] New post 22 Oct 2005, 08:49
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 21 Jun 2015, 10:07, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: In how many ways can 5 identical black balls and 7 identical [#permalink] New post 22 Oct 2005, 11:18
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fresinha12 wrote:
8C3? or 8C5?

gsr wrote:
A) 8C3 = 56


8C3 = 8C5 :!:
Position of white balls is fixed. Since they are identical, number of ways to arrange them is 1.
We have 6 spaces in btw the white balls and 1 each to the left and right end to accomodate the black ball. So 8 places in total to arrange 5 balls
8C5 (or 8C3)
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Re: In how many ways can 5 identical black balls and 7 identical [#permalink] New post 23 Oct 2005, 07:24
there are 8 spaces availale so that no two black balls are together. that means 8! arrangements. but there are 5 identical black balls and 3 identical spaces. that`s why we have to divide by 5! and 3!. 8!/(5!3!).
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Re: In how many ways can 5 identical black balls and 7 identical [#permalink] New post 21 Jun 2015, 06:39
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Re: In how many ways can 5 identical black balls and 7 identical [#permalink] New post 21 Jun 2015, 08:18
Can anyone help me with this? I have no idea how to solve this.
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Re: In how many ways can 5 identical black balls and 7 identical [#permalink] New post 21 Jun 2015, 10:37
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roopika2990 wrote:
Can anyone help me with this? I have no idea how to solve this.


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

Consider the following arrangement:

*W*W*W*W*W*W*W*

Now, if we replace any 5 out of 8 stars with black balls, then no 2 black balls will be together.

# of ways to choose 5 out of 8 is \(C^5_8=56\).

Answer: A.

Check other Seating Arrangements in a Row and around a Table questions in our Special Questions Directory.
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In how many ways can 5 identical black balls and 7 identical [#permalink] New post 21 Jun 2015, 10:56
roopika2990 wrote:
Can anyone help me with this? I have no idea how to solve this.


Consider the possible arrangements:

WBWBWBWBWBWW or BWBWBWBWBWWW or WWBWBWBWBWBW or WWWBWBWBWBWB

Clearly, we are concerned about not placing 2 B's together, whites can be placed however we want.

So, it is like _W_W_W_W_W_W_W_ where black ball can be placed anywhere on the blank spaces, and this way, there never will be a case where 2 black balls fall together.

As there are 8 black spaces where 5 black balls can be placed, thus, the combination turns out as 8C5 (again, we are not concerned about the placement of white balls as we have already "fixed" their positions in the diagram above)

Also, we achieved at 8C5 as arranging 5 black balls on 8 different spots => 8!/5!3! (Total combination of available spaces = 8!; choosing 5 spaces out of them = 5!, arranging 3 leftover spaces = 3!) = 56

Hence answer is A
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Re: In how many ways can 5 identical black balls and 7 identical [#permalink] New post 27 Jul 2015, 09:07
King407 wrote:
roopika2990 wrote:
Can anyone help me with this? I have no idea how to solve this.


Consider the possible arrangements:

WBWBWBWBWBWW or BWBWBWBWBWWW or WWBWBWBWBWBW or WWWBWBWBWBWB

Clearly, we are concerned about not placing 2 B's together, whites can be placed however we want.

So, it is like _W_W_W_W_W_W_W_ where black ball can be placed anywhere on the blank spaces, and this way, there never will be a case where 2 black balls fall together.

As there are 8 black spaces where 5 black balls can be placed, thus, the combination turns out as 8C5 (again, we are not concerned about the placement of white balls as we have already "fixed" their positions in the diagram above)

Also, we achieved at 8C5 as arranging 5 black balls on 8 different spots => 8!/5!3! (Total combination of available spaces = 8!; choosing 5 spaces out of them = 5!, arranging 3 leftover spaces = 3!) = 56

Hence answer is A


Hi King407,
Can you elaborate how there is 8 places although there are 5 black balls? I do not get it

Thanks
Re: In how many ways can 5 identical black balls and 7 identical   [#permalink] 27 Jul 2015, 09:07
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