In how many ways can we distribute 10 identical ice creams to 4 people so one of them gets a maximum of 4 ice cream?

Question

I know this formula should be used but I do not know how to apply the maximum part $\binom{n+r−1}{r−1}$

Edit: Possible answers are: $\binom{8}{5}$, $\binom{13}{10}$-$\binom{8}{5}$,$\binom{10}{4}$-$\binom{7}{4}$,$\binom{13}{4}$-$\binom{5}{4}$ or $\binom{10}{4}$-$\binom{6}{3}$

Edit2: since there are identical ice creams, some people may get 0 while one may even get 10. But one of them can get MAXIMUM 4.

Answer 1

I believe using principle of inclusion and exclusion (PIE) is the easiest.

if nobody gets more than 4 ice creams:

$$ \begin{aligned} \sum_{i=0}^{\left\lfloor\frac{10}{5}\right\rfloor}{\binom{4}{i}\binom{13-5i}{3}\left( -1\right)^{i}}&=\binom{4}{0}\binom{13}{3}-\binom{4}{1}\binom{8}{3}+\binom{4}{2}\binom{3}{3}\\ &=68 \end{aligned} $$

if only one particular person cannot get more than 4 ice creams:

Say that person A cannot have more than 4. We want to count the number of distribution subtracted by number of distribution in which A get more than 4.

$$ \binom{13}{3}-\binom{8}{3}=230 $$

here, $\binom{8}{3}$ means we distribute 5 candies to 4 people then add 5 candies to person A. this way he always get more than 4.

You may want to read about this PIE, a very powerful tools in combinatorics.

Answer 2

Good evening,

First, you have to decide which one of the people gets the maximum of 4 ice creams. You have $ p=4 $ possibilities for that. Therefore, you then just have to choose how to spread the 6 remaining ice creams among the 3 other people. This comes down to choose the places of the sticks "|" in the following expression :

            OO|O|OOO  (first has 2, second has 1 and third has 3) 

For each stick, you have 7 different positions possible, so in total 7*7=49 possible configurations. Then, you need to remove the permutations induced by the fact that the sticks are non-differentiable : stick 1 in position 3 and stick 2 in position 5 is the same than stick 1 in position 5 and stick 2 in position 3. To remove those cases, we can divide by 2!, the number of permutations of the sticks.

In the end, since you had 4 possibilities for the guy getting the 4 ice creams, this leads to (49/2)*4=98 choices.

Hope this helps,

**Edit : ** my misstake, are you looking for :

• the number of cases where one get exactly 4 ice creams ?

• the numer of cases when one of the guys gets max 4 ice creams ?

• the number of cases where any of the guys gets max 4 ice creams ?

In all cases, it can be done with the previous technique I gave you, by counting different cases.

Edit2 : Corrected the number of permutations with the sticks according to comments (thanks ! ). And it seems this case corresponds to what has been asked.

Answer 3

An alternative can be:

Name the people $a,b,c,d$

Now we want solutions to the equation

$$a+b+c+d=10$$ where $0\leq a,b,c,d \leq 4$.

This is equivalent to finding the coefficient of $x^{10}$ in the following expression:

$(1+x+x^2+x^3+x^4)^4$

The expansion can sometimes get ugly but this is better when there are restrictions for which there are many subcases.

Answer 4

Since there are $4$ people and $10$ ice creams, someone must get at least $3$, so the maximum anyone gets is either $3$ or $4$. The possible patterns when the maximum is $3$ are $$3,3,3,1\\3,3,2,2$$ The possible patterns when the maximum is $4$ are $$4,4,2,0\\4,4,1,1\\4,3,3,0\\4,3,2,1\\4,2,2,2$$ Now you just have to figure out how many ways each of these patterns can arise and add them up.

Usually, there are too many cases fo this to be a practical procedure, but I think it's the best way for this problem.