In how many different ways can all of 5 identical balls be placed

Question

https://gmatclub.com/forum/download/file.php?id=34040 In how many different ways can all of 5 identical balls be placed in the cells shown in the figure such that each row contains at least 1 ball? (A) 64 (B) 81 (C) 84 (D) 96 (E) 108 Figure 1.JPG

chetan2u · Accepted Answer

Hi,

Another way..

1) Total ways of selecting 5 places out of 9 = 9C5= \frac{9!}{5!4!}=126 ..

2) ways in which the 5 places are in just two rows..
 Selecting 2 rows out of 3=3C2=3..
 These two rows have 6 cells, so choosing 5 out of 6 cells=6C5=6..
So total ways in which the balls are NOT in all three= 3*6=18..

Ways in which the balls are in all three cells=126-18=108

E

Kurtosis · Answer

The balls can be placed in the rows in 2 ways --> 3 - 1 - 1 or 2 - 2 - 1

Number of different ways = (3C3*3C1*3C1)*3 + (3C2*3C2*3C1)*3 = 27 + 81 = 108 ways

Answer: E

AR15J · Answer

It is not mentioned in the ques that one cell can have only one ball. Both the solution mentioned above are based on this assumption. How did we assume it? Please help.. This assumptions are killing me

niks18 · Answer

hi  AR15J

it is clear from the image that there are 3 rows. It is mentioned in the question that each row has at least 1 ball. So you have 5 balls to play with and 3 rows how are you going to distribute them?

AR15J · Answer

Thanks Niks for your reply.. I understand this.. But why can't we place more than one ball in a cell. For example , to calculate -in how many ways 2 balls can be place in a row.. why the no of ways are -- 3c1*3c2? (selecting a row and then selecting two columns from that row to place two balls)

Why are we not considering the case when both the ball are in the same cell of a row? If we consider this, total cases will be 3c1*3c2+3c1

niks18 · Answer

Hi AR15J ohhhk now I got your point. So ideally a standard GMAT official question would have stated clearly that 1 cell contains 1 ball only. The solutions above have also assumed that 1 cell contains 1 ball only. As this is a question from an unknown source, i would suggest let's not think too much into that

Hi Bunuel what are your thoughts about the language of the question?

testcracker · Answer

what is more, it is stated that EACH ROW NOT EACH CELL must contain at least 1 ball so we can first place 1 ball in each row, as there are 3 rows, we are left with (5 - 3) = 2 balls now, as the rows and balls are identical, we can place the 2 balls in 3 rows as under 4! ____ 2!2! = 6 ????

jim441 · Answer

There are three throws and each row must have atleast one ball.
So, there are two cases:

1. One row has 3 balls, other two rows have one ball each.
2. Two rows have 2 balls each and one of the row has one ball.

so for case 1: 
Row with three balls: 1 possible way
Row with one ball: 3C1 (selecting one of three cells of the row)
Row with one ball: 3C1 (selecting one of three cells of the row)
and now arrangement of the rows: 3!/2!

case 1 = 1*3C1*3C1*3!/2! = 27

case 2:
Row with one balls: 3C1
Row with two ball: 3C2 (selecting two of three cells of the row)
Row with two ball: 3C2 (selecting two of three cells of the row)
and now arrangement of the rows: 3!/2!

case 2 = 3C1*3C2*3C2*3!/2! = 81

so total cases = 27+81 = 108

DhanyaAbhirami · Answer

Is there a way to assign 1 ball to each row and then assign remaining 2 balls ?

aneebunnee · Answer

I used the following approach and got a completely wrong answer. Could someone help me understand where I'm going wrong?

As there must be one ball in each row, I assumed we can first place one ball in each in three different ways:
so 3 x 3 x 3 = 27

Then, as there are 6 spots remaining with 2 balls, can we not do 6C2? But that gives an answer of 15, and multiplying that with 27 obviously gives a far too big answer.

Where did I go wrong in my thought process?

Bunuel · Answer

For example, if with 3 * 3 * 3 you place a ball in the first cell of each row and then use 6C2 to place the remaining two balls in the last two cells of the first two rows, we get the following case: GMAT-Club-Forum-3zpynmsd.png However, if with 3 * 3 * 3 you place a ball in the last two cells of the first two rows and the first cell of the third row, and then use 6C2 to place the remaining two balls in the first cells of the first two rows, we get the following case: GMAT-Club-Forum-864xfmtp.png As we can see, these two cases are identical, leading to overcounting.

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 different ways can all of 5 identical balls be placed

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 different ways can all of 5 identical balls be placed

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