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A shop offers infinite number of 7 distinct flavours of ice cream. In how many ways can the customer create his/her portion given that order can consist of at most 5 scoops of ice cream and that the case of cup without ice creams is neglected?

Made corrections to my initial solution.

Solution:

let $A_{i}$ -multiset of size $i$ of all possible ice cream portions customer can order ($i\in \{1,2,3,4,5\}$). Consider computation of $|A_{i}|$, except $|A_{1}|$, which is trivially equal to 7, by the use of "stars & bars "method. There are 8 bars for, $i$ stars, giving directly combinations with repetitions size $i$ from the set consisting of $7+i-1 = 6 + i$ elements. Hence: $|A_{i}| ={{6+i}\choose{i}} = \frac{(6+i)!}{i! 6!} $. Also: $ (i\neq j) \implies A_{i} \cap A_{j}=\emptyset$. The rule of sum implies: number |S| of all possible orders is: $|S|=|\bigcup\limits_{i=1}^{5}A_{i}|=\sum\limits_{i=1}^{5} |A_{i}| =\sum\limits_{i=1}^{5}\frac{(6+i)!}{i!6!}=\frac{7!}{1!6!} + \frac{8!}{2!6!} + \frac{9!}{3!6!} + \frac{10!}{4!6!} + \frac{11!}{5!6!} = 7 + 28+\frac{9 \times 8 \times 7}{3\times2} + \frac{10\times9\times8\times7}{4\times3\times2} + \frac{11\times10\times9\times8\times7}{5\times4\times3\times2}=7+28+84+210+462 = 791 $

Can anyone state whether my current solution is correct? Thanks for help in advance!

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    You can immediately see that the solution you propose could not be correct since your answer is not even an integer. The answer to any counting question will always be an integer. – 2017-02-28
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    Since no one has said, I'll say it: Kudos to you for attempting the problem and asking for help, rather than just asking us to do it. We appreciate that. +1 – 2017-02-28
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    Posted improved version of my solution. Could anyone give feedback? – 2017-03-01
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    The way you have it set up $A_i$ is the set of all orders of size $i$, where each order is a multiset of size $i$. Maybe edit to reflect that? – 2017-03-01
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    **(1)** You have introduced an auxiliary $8^{th}$ flavour but have not used it anywhere. **(2)** If you have $8$ choices in selecting 1 scoop, $A_1$ is trivially $8$, not $7$. **(3)** Having $8$ bars means you have $9$ types, which is not true. **(4)** Surely $ \binom{7+i}{i} \neq \frac{(6+i)!}{i! 6!}$ **(5)** I suggest you go over the argument and rewrite the proof from scratch so as to understand it better and avoid confusing yourself. – 2017-03-01

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No, you can't just divide by $i!$, because not all combinations are counted that many times -- for example for a 4-scoop order, $7^4$ counts "vanilla, vanilla, vanilla, vanilla" only once.

(A different hint that something is awry is that the only prime that divides $7^i$ is $7$, so when $i\ge 2$, $7^i/i!$ is not even an integer!)

A better strategy: Add an 8th flavor, "nothing", and use stars-and-bars to figure out the number of ways to write 5 as the sum of 8 non-negative integers. Then subtract one for the "5 × nothing" order.

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    Thanks for feedback. I overcomplicated this problem. It could be solved direclty using "stars & bars" method considering each size of order separately. – 2017-03-01
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You've correctly figured out that $i \neq j \implies A_i \cap A_j = \emptyset$; that's a good start!

The last part is counting what each $A_i$ contains. [Since Henning Makholm has already pointed out the errors I shall only provide an alternative method of counting.]

Converting the problem into our trusty stars and bars sequence, you get $i$ ice creams(stars) separated into $7$ categories by $6$ bars. If you have $x$ stars appearing before the first bar, then that represents $x$ ice creams of the first type; if $x$ stars appear between the first and second bar, it represents $x$ ice creams of the second type, and so on. So, count the number of sequences of $i$ stars and $6$ bars.

This is the usual way of counting $i$-sized multisets whose elements all belong to a given set, which in this case is $[7]$.

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    Here is the set of all orders, $O=\{\{1\},\{2\},...\{1,2\},\{2,3\},...\{1,2,3\},\{1,1,2\}...\}$. The set of all orders of $2$ scoops $A_2 =\{\{1,2\},\{2,3\}...\{3,3\}\}$ - note that each element is a set of size $2$, so when you take intersection with all orders of size $4$ - $A_4$, you get the empty set. – 2017-03-01
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    Correction: note that each element is a **multiset** of size $2$. – 2017-03-01
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    Posted improved version. – 2017-03-01
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    @MIT Your arithmetic is wrong, 3rd term is $84$, not $168$. Check the rest too. $\sum_{i=1}^{5} \binom{i+6}{6} = 791$ is the correct value. – 2017-03-01
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    Correction implemented in my solution. Checked other computation. Now final result is 791, the same as yours. Is now my solution, including computations, correct? – 2017-03-01
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    Yes, the final answer is correct but the counting argument seems muddled; please read my suggestion in the comments to your question. – 2017-03-01