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

Question

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

Bunuel · Accepted Answer

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 .

gsr · Answer

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)

christoph · Answer

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!).

roopika2990 · Answer

Can anyone help me with this? I have no idea how to solve this.

King407 · Answer

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

Mo2men · Answer

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

Thanks

sairam595 · Answer

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

chetan2u · Answer

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

kham71 · Answer

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

Radhika11 · Answer

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!

OreoShake · Answer

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.

Bunuel · Answer

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.

hatemnag · Answer

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.

arirux92 · Answer

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

Kinshook · Answer

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

TARAUJO · Answer

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?

Regor60 · Answer

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

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards