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

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!).

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

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

Here's a different way to solve it. Posted an incorrect response earlier but eventually figured it out.

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!

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.

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 hatemnag

Let 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 Cr-1 = (3 + 6-1)C(6-1) = 8C5 = 56

If you have any doubt about the formula pm me, I will clarify the formula for you

Asked: 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

_BW_BW_BW_BW_B_

Out of 7 white balls, 4 white balls are to be placed in between 2 black balls surrounding them. Remaining 3 white balls can be placed in 6 places available.

Case 1: All 3 white balls are together.
6 ways

Case 2: 2 white balls are together, 1 white balls is seperate
6*5 = 30 ways

Case 3: No white balls are together
6*5*4/3! = 20 ways

Total ways = 6 + 30 + 20 = 56

IMO A

Can someone please advise if the way I solved is correct or if my method simply gave me the correct answer by chance.

How I tackled it was:

12C5 (choosing 5 black balls out of the 12 total)

12C5=72

Then, for the black balls I thought of it as, you do not want to choose two black balls successively so as to not have two black balls directly next to each other; therefore, 5C2 and 4C2 (probability of choosing two balls successively out of 5 and then out of 4)

5C2= 10
4C2=6

From 12C5 I subtracted the possibility of choosing two black balls successively (5C2 and 4C2)
72-10-6=56

Does this method make sense or was it just mere coincidence that it worked out?

This is an arrangement problem not a selection problem. All balls have to be distributed.

The combination 12!/7!5! is not selecting the 5 black balls as you suggest. It's selecting any 5 of the 12 balls.

If you think about it, there is only 1 way to select 5 black balls, because there are only 5 to be selected.

So, it may be a coincidence

Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.