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Re: In how many ways can 5 identical black balls and 7 identical [#permalink]

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22 Oct 2005, 12:18

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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)

Re: In how many ways can 5 identical black balls and 7 identical [#permalink]

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23 Oct 2005, 08:24

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]

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21 Jun 2015, 07:39

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

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21 Jun 2015, 11:56

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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

Re: In how many ways can 5 identical black balls and 7 identical [#permalink]

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27 Jul 2015, 10:07

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

don't we consider here arranging 7 white balls in 7 places as 7!

so that num of ways can 5 identical black balls and 7 identical white balls be arranged in a row so that no 2 black balls are together is 7!*56
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don't we consider here arranging 7 white balls in 7 places as 7!

so that num of ways can 5 identical black balls and 7 identical white balls be arranged in a row so that no 2 black balls are together is 7!*56

Hi, since the balls are identical, all 7 balls can be arranged at 7 places only in one way... yes had the ball been not identical, there would have been 7! ways
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Re: In how many ways can 5 identical black balls and 7 identical [#permalink]

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12 Apr 2016, 05:53

Hi,

Is it correct to say in this question that the arrangement of balls within their group does not matter in this question as the white balls and black balls are identical within their respective groups?

Also is there a way to answer is by the 'reduction from all possible arrangements' method? i.e 12! - something? I thought that could be possible as there is condition for the black balls to not be togher; hence we can maybe 'glue' them and reduce them from all possible arrangements. Not sure how to do that though.

Is it correct to say in this question that the arrangement of balls within their group does not matter in this question as the white balls and black balls are identical within their respective groups?

Also is there a way to answer is by the 'reduction from all possible arrangements' method? i.e 12! - something? I thought that could be possible as there is condition for the black balls to not be togher; hence we can maybe 'glue' them and reduce them from all possible arrangements. Not sure how to do that though.

We are told that 5 black balls and 7 white are identical, so we are not bothered about their arrangements. As for your second question, yes, it's possible to do it with (total) - (restriction) approach but it will be much more tedious than the approach used above.
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