Thread: A box contains 36 identical sweets

1. A box contains 36 identical sweets

How many different ways can the sweets be divided among 6 kids so that each of them receives atleast 3 sweets?

Firstly I'd give each of them 3 sweets, a total of 18.

Then distribute the other 18 any way possible? Not sure how to go about that though?

Thanks in advance for any help.

2. You need to count the number of ways to represent 18 as a sum of 6 natural numbers (possibly equal to zero). See this Wiki page.

3. The number of ways the put N identical objects into K different cells is:
$\dbinom{N+K-1}{N}.$

4. Originally Posted by Plato
The number of ways the put N identical objects into K different cells is:
$\dbinom{N+K-1}{N}.$
Ok I did something similar. Here's what I got:

If you had to find the number of different ways that 36 identical sweets can be divided among 6 kids it would 41Choose5 so the answer is 749,398.

Ofcourse I want each of the kids to have atleast 3 sweets. So I'd say the answer to my original question is 23Choose5 = 33,649.

Is that right?

5. $\dbinom{18+6-1}{18}$
You have already used 18, three to each child.
You have 18 left. They are identical.
There are still 6 different children.

6. Originally Posted by Plato
$\dbinom{18+6-1}{18}$
You have already used 18, three to each child.
You have 18 left. They are identical.
There are still 6 different children.
and that is the same as 23choose5 which is 33,649.

7. Hello, Markhor!

How many different ways can 36 sweets be divided among 6 kids
so that each of them receives atl east 3 sweets?

Firstly I'd give each of them 3 sweets, a total of 18.

Then distribute the other 18 any way possible? . Yes!

We will distribute the other 18 sweets among the 6 kids.

For each sweet, we have 6 choices of kids to give it to.

So there are: . $6^{18}$ possible distributions.

. .
Edit: This is wrong! . . . *blush*

There are: . $19^5$ distributions.

8. Originally Posted by Soroban
So there are: . $6^{18}$ possible distributions.
It is reasonable to assume that children are numbered but sweets are not. In fact, the OP says that the sweets are identical.

Now, $6^{18}$ is the set of all 18-tuples $(x_1, ..., x_{18})$ where $1\le x_i\le 6$ for each $i=1,\dots,18$. Each $x_i$ says to which kid the $i$th sweet goes. However, since sweets are identical, we don't care whether sweets #1 through 9 go to kid #1 and those #10 through 18 go to kid #2 or vice versa. So, in our model we count not the number of ordered tuples, but the number of multisets $\{x_1,\dots,x_{18}\}$. This is $\left(\!\tbinom{6}{18}\!\right)=\tbinom{23}{18}=\t binom{23}{5}$.