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In how many ways can 5 identical black balls and 7 identical [#permalink]
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Updated on: 21 Jun 2015, 11:07
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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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Originally posted by FN on 22 Oct 2005, 09:49.
Last edited by Bunuel on 21 Jun 2015, 11:07, 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]
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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)



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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, 09:18
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]
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21 Jun 2015, 11:37
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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 Hence answer is A



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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 Thanks



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Re: In how many ways can 5 identical black balls and 7 identical [#permalink]
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27 Dec 2015, 02:36
Bunuel wrote: 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. Hi Bunuel, 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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Re: In how many ways can 5 identical black balls and 7 identical [#permalink]
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27 Dec 2015, 04:35
smartguy595 wrote: Bunuel wrote: 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. Hi Bunuel, 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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15 Jan 2016, 13:33
Here's a different way to solve it. Posted an incorrect response earlier but eventually figured it out.
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Re: In how many ways can 5 identical black balls and 7 identical [#permalink]
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26 Mar 2016, 09:45
Bunuel wrote: 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. Hi Bunnel Can you please help me understand why can't we solve it using Total arrangements  Number of arrangements where blacck balls are always together Total = 12!/(7!*5!) Black balls together: 8!/7!



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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.



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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, 06:55
abypatra wrote: 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. 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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Re: In how many ways can 5 identical black balls and 7 identical [#permalink]
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20 Apr 2016, 05:19
Hi Bunuel
why we should start arrangement with black not white. i guess if we start with white, it would be 7 black places. please help.



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Re: In how many ways can 5 identical black balls and 7 identical [#permalink]
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19 Jul 2016, 05:21
hatemnag wrote: Hi Bunuel
why we should start arrangement with black not white. i guess if we start with white, it would be 7 black places. please help. Hi hatemnagLet me show you the way with black balls fixed B B B B B Now restriction is that no two black ball can be together So 4 white balls fixed B W B W B W B W B Now 3 white balls remain. But remember, all are identical so it can't be 18C3 So spaces remaining _ (BW) _ (BW) _ (BW) _ (BW) _ B _ = 6 spaces Where each blank space can have 0 to 3 balls each Now we have to divide 3 identical balls among 6 spaces, where each can be from 0 to 3 I hope you know the formula for dividing identical objects among groups : n+ r 1 Cr1 = (3 + 61)C(61) = 8C5 = 56 If you have any doubt about the formula pm me, I will clarify the formula for you
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Re: In how many ways can 5 identical black balls and 7 identical [#permalink]
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