Five balls of different colors are to be placed in three different box

Another way:

\(N = 3^5 - 3*2^5 + 3 = 243 - 96 + 3 = 150\)

\(3^5\) - the total number of ways to put 5 different balls into 3 different boxes. Each ball has 3 possible options (boxes)

\(3*2^5 - 3\) - the total number of ways to put 5 different balls into 1 or 2 different boxes out of 3 (to leave at least 1 box empty).

Could you please break down the highlighted part and explain how are you getting those figures.....

If there would not have been boxes, 5 different balls we could arrange in 5! = 120 ways.
Now look at a given arrangement of the 5 balls, let's say ABCDE. Because we have three different boxes, then the following arrangements are possible:

|AB| |CD| |E|; |AB| |C| |DE|; |A| |BC| |DE| - these are all 2,2,1 type scenarios

and they are considered different arrangements (this would give 5! * 3 possibilities), EXCEPT THAT, inside each box, if we have more than one ball, order doesn't matter. So, when in a box we have for example AB, this is the same as BA, we don't mind whether A or B was put first into the box. Therefore, we should divide by k! whenever we have in a box k balls. 1! doesn't change the value of the expression, I left them just for the completeness of the formula.

Similar reasoning applies for the 3,1,1 scenario.

Eva,
I was able to compute the combinations for distributing balls in 2-2-1 and 3-1-1 way. However, I used 3! for the arrangement of the boxes. Let's say that the boxes are labeled X Y and Z -- and we are using (2,2,1) type of placement of balls. Wouldn't the order of boxes matter? i.e. 3!? That is, I could have any of the three boxes in the first spot, 2 for the second spot and 1 for the third one?

Example : abcde are balls and three boxes are xyz =>
X=[AB] Y=[cd] Z=[e]
X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6

I am a bit confused. Still thinking....Please help me. thanks

You keep the boxes fixed, you permute the balls.
For example, if you have in the boxes |AB| |CD| |E|. This is obtained from ABCDE.
You also have |CD| |E| |AB|, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls.

For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities:
|AB| |CD| |E|
|AB| |C| |DE|
|A| |BC| |DE|

So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example:
AB in the first box is the same as BA, and CD in the second box is the same as DC.
All the arrangements
|AB| |CD| |E|
|BA| |CD| |E|
|AB| |DC| |E|
|BA| |DC| |E|
should be counted just once.

Therefore, 5!*3/(2! * 2! * 1!).

I did it the way walker did since we can avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small). Do visit that solution properly too. In this question, consider it an exercise in lots of PnC concepts.

In how many ways can you distribute 5 different balls in 3 different boxes such that a box may get no ball? (we will later subtract those cases in which one or two boxes are empty)
Each ball can be placed in any of the 3 boxes so there are 3 ways to place a ball.
Since there are 5 balls, we can place all the balls in 3*3*3*3*3 = 3^5 ways

Only one box empty:
In how many ways can you distribute the 5 different balls in 2 different boxes such that both boxes have at least one ball? (we want to find the cases in which one and only one box will be empty)
Select the empty box in 3 ways. Each ball can be placed in any of the 2 boxes. So 5 balls can be placed in 2*2*2*2*2 = 2^5 ways. Out of 2^5, subtract those cases where the balls are all in 1 box only. This happens in 2 ways since you can select an empty box again out of the two in 2 ways. Total number of ways = 3*(2^5 - 2).

2 boxes empty:
In how many ways can you distribute the 5 balls in only 1 box? In 3 ways since you select a box in 3 ways.

You can distribute 5 balls in 3 boxes such that no box is empty in 3^5 - 3*(2^5 - 2) - 3 = 243 - 90 - 3 = 150 ways

For a discussion on distribution of objects, check out these posts too:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/12 ... 93-part-1/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/12 ... s-part-ii/

This is the beauty of combinatorics: doesn't matter how you execute your plan (here, that of distributing some balls in boxes), if the stages are correctly translated into mathematical expressions, you always get the correct answer. We have already seen three different approaches to solve the question.
Just for the fun, I tried to reach the correct answer with another scenario:

Two stages:
1) Distribute one ball in each box, to be sure that there is at least one ball in each box. This can be done in 5*4*3 = 60 ways, boxes are distinct, order matters.
Continue with stage:
2) Place the remaining 2 balls.
Here we have two possibilities:
2a) Either both balls will be placed in one of the 3 boxes - OR -
2b) Each ball will be placed in a separate box.

For 1 and 2a) - 60*3/3 = 60; 3 possibilities to choose the box that will get the remaining 2 balls; divide by 3, because in the box with the 3 balls, it doesn't matter which ball was placed in at stage 1, it could be any of the final 3 balls.

For 1 and 2b) - 60*3*2/4 = 90; 3 possibilities to choose the box that will not get an extra ball, which is equivalent to choosing 2 boxes that will each get an extra ball; 2 possibilities to place the remaining balls in the 2 different boxes; divide by 4 = 2*2, because in those 2 boxes with 2 balls, it doesn't matter in which order the 2 balls arrived, stage 1 or stage 2b).

Again, we obtain the total of 60 + 90 = 150 possibilities.

Eva,
Thanks for your help. I think that I see your point. Do you think that if the boxes are placed on a rack in a store (order of the three boxes will matter -correct?), then we will have to multiply by 3!? That is, for 2-2-1 case, 5C2*3C2*1*(3!) ---Please let me know your thoughts....

{ My approach:
Here's how I approached this problem :- for 2-2-1 case, the combinations will be 5C2*3C2*1C1 (choose 2 balls from 5 balls - then - choose 2 balls from the remaining 3 balls - then - choose 1 ball from 1 ball)

Now, since the boxes are labeled differently, we need to permute them. I initially thought of multiplying by 3! because of these combinations :
X[AB] Y[cd] Z[e]
X[AB] Z[e] Y[cd]
X[CD] Y[AB] Z[e]
X[CD] Z[e] Y[AB]
X[e] Y[AB] Z[CD]
X[e] Z[CD] Y[AB]
........ditto for 3-1-1 case.
}

I am a bit confused.


Thanks

You don't permute the boxes! Once the boxes are defined, they are unique. You called them X, Y, Z. Place them on a shelf, nailed them down, don't touch them!!! The question is not about arranging the boxes on a shelf, it's about their contents. Even if you would arrange the boxes in a different order, say Y, X, Z, what matters is only their content and not the order in which they are displayed.

For the combinatorial approach, see Bunuel and Walker's solutions.

Here, I was playing a different game. I was permuting the balls and then decide (define) the boxes.

Try to understand each solution separately, and don't mix between them.

Just a remark:

...avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small).

There is no need for enumeration, I wrote them down because the list is indeed short. For each scenario, there are 3C1=3, 3C2=3, possibilities, respectively.

Can someone please explain the text I have marked in Red?

You have 3 boxes. YOu want to keep exactly one empty. This means the other two must have at least one ball. Say the boxes are A, B and C. In how many ways can you choose to have exactly one box empty?
A empty, B and C non empty
B empty, A and C non empty
C empty, A and B non empty

Basically you can select the box you will keep empty in 3 ways.

Now you have to put 5 balls in 2 boxes (say e.g. B and C) such that both boxes have at least one ball.

For each ball, there are two options: box B or box C
So you can put in 5 balls in 2*2*2*2*2 = 2^5 ways.
But this includes cases where all ball are put in one box only. How many such cases are there? All balls could go in box B and box C could be empty. All balls could go in box C and box B would be empty. Only these two cases are possible. So out of the 2^5 cases, you need to remove these two cases because you must have at least one ball in each of the two boxes, B and C.

No of ways in which you have all balls in exactly two boxes = 3*(2^5 - 2)

How many ways can you distribute 5 marbles in 3 identical baskets such that each basket gets at least 1 marble.

I am trying to solve this problems using as many approaches:

The correct answer is:
we can pick distribute the marbles in two such ways:
3-1-1 or 2-2-1.
For the 3-1-1. We choose 3 marbles for the 1st basket. (5C3). 1 for the 2nd basket (2C1) and from the remaining marble (1C1).
We then divide by 2!. I know it has something to do with the the fact we are putting 1 marble into the 2 baskets but what is the intuition?? How do you if you have to divide by n!?? This is what I am not getting. After we have (5C3)(2C1)(1C1) does the problem become... " How many ways can we rearrange the digits 311?" Ans: 3!/(2!1!)

If I choose 3 people out of 5 for a soccer game its 5*4*3/3!. I divide by 3! Because order does not matter. How does this translate to the above problem?

Hi,

You see the constraint that each basket has to have at least 1 marble. So put 1 marble in each basket. We have 2 marbles remaining. This can be distibuted among 2 baskets or put in just 1 basket. The former can be done in 3C2 ways and the latter in 3C1 ways. So there are totally 6 possible ways you can do this.

I assume the marbles are identical. So you do not have to consider 5C3.

Hi i did it that way and i don't understand why i'm wrong:

We can have two distributions:

1. 3-1-1 one box gets three balls, the 2 other boxes only 1

It is equivalent as to make 3 groups ordered out of 5 objects (as there are 3 different boxes)

So 5C3 * 2C1 * 1C1 = 20

2. 1-2-2 one box gets one ball and the remaining two boxes get two balls each.

It is equivalent as to make 3 groups ordered out of 5 objects (as there are 3 different boxes)

So 5C1 * 4C2 * 2C2 = 30

so 50 solutions..

The boxes are different B1, B2, B3
When you split the balls 3-1-1, which box gets 3 balls? You first select a box which gets 3 balls. You can do this in 3 ways - select B1 or B2 or B3.
Now select 3 of the 5 balls for this box in 5C3 ways. Now you have 2 balls and 2 boxes one for each ball so you can distribute them in 2 ways.
This gives 3*5C3*2 = 60.

Same problem for 2-2-1 case. You multiply by another 3 and you get 3*30 = 90

Total 60+90 = 150

when can we use (n-1)C(r-1) ?
It is equal to number of ways of dividing n things tor people where each can receive atleast 1? isnt it the case here?