# Permutations and Combinations: Distributing Balls Among People

• erisedk
In summary, the total number of ways to distribute 5 different colored balls among 3 people so that each person receives at least one ball is 150. This can be calculated by first choosing the person who will receive 3 balls, then assigning the remaining 2 balls to the other 2 people. Alternatively, this can be calculated by finding the number of ways to give 2 people 1 ball each and giving the remaining ball to the third person. The mistake in the calculation was counting different orders of the persons receiving the balls as different distributions.
erisedk

## Homework Statement

The total number of ways in which 5 balls of different colors can be distributed among 3 persons so that each person gets at least one ball is
Ans: 150

## The Attempt at a Solution

I don't understand what's wrong with my answer.
In case of each person getting one ball and the remaining two balls going to a single person, I have 5*4*3*3 ways. 5*4*3 for the first three balls and then the remaining two balls have three options, therefore the final *3

In case of the remaining two balls going to different people, I have 5*4*3*6 ways to do so, the final*6 because I have 3 spots and I have two pick two balls, and order matters as the balls are different.

Adding 5*4*3*3 + 5*4*3*6, I get 540.

What am I doing wrong?

Last edited:
50 is wrong, but 540 is not right either. You also count the order the persons get the balls as different. If the colors are ABCDE and the persons are 1, 2, 3, you count "A to 1, B to 2, C to 3 and then D and E to 1" as different from "D to 1, B to 2, C to 3 and then A and E to 1".

Better start picking the person to get 3 balls, and then look at the other two persons. The other part has the same issue.

Another way to look the 1-1-3 distribution is is to count the number of ways you can give two of the three people one ball each. The third person ends up with the rest.

mfb said:
You also count the order the persons get the balls as different.
erisedk, just in case that's not clear, mfb is saying you mistakenly counted those orders as different, not that you should count them as different.

Oh, ok.
I get it now.
Thanks :)

## 1. What is the difference between permutations and combinations?

Permutations are ordered arrangements of a set of items, while combinations are unordered selections of items from a set.

## 2. How do I calculate the number of permutations or combinations?

For permutations, use the formula n!/(n-r)! where n is the total number of items and r is the number of items in each arrangement. For combinations, use the formula n!/r!(n-r)! where n is the total number of items and r is the number of items in each combination.

## 3. Can I use permutations and combinations in real life situations?

Yes, permutations and combinations are used in various fields such as mathematics, computer science, finance, and statistics. Examples include calculating the number of possible outcomes in a game of chance, determining the number of possible combinations of ingredients in a recipe, and analyzing data in research studies.

## 4. What is the difference between with and without replacement in permutations and combinations?

With replacement means that an item can be chosen more than once in a permutation or combination, while without replacement means that each item can only be chosen once.

## 5. Can permutations and combinations be used in combination with other mathematical concepts?

Yes, permutations and combinations can be used in combination with other concepts such as probability and counting principles to solve more complex problems. They are also used in the study of combinatorics, which deals with counting and arranging objects.

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