Some of us already did, and some of us are going to end the years with the word "teen" in it soon, for another 94 years.

So let me ask the question: How many distinct numbers can you produce with using the numbers
10, 9, 8, 7, 6, 5, 4, 3, 2, 1
in this exact order, by using only addition, subtraction, multiplication, division, and brackets?

An example is
10 x 9 x 8 x (7 + 6) x 5 x 4 x (3 + 2) x 1 = 2020.
A point for each number, and two points for a 2020 as a result.

  • $\begingroup$ Is this supposed to be a combinatorics problem or an open-ended, points based game? $\endgroup$
    – Adam
    Dec 31, 2019 at 19:04

3 Answers 3





Partial answer with quick upper and lower bounds for the range.

I believe the highest number you can get should be:

$10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times (2 + 1) = 5\ 443\ 200$

The lowest should be:

$10 - (9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times (2 + 1)) = -544\ 310$

Which means that at most there could be:

$6\ 987\ 511$ distinct integers. The extra one is for zero.

That said, the problem states numbers, which means we need to include decimals.

Heading out shortly, but thoughts about an answer that probably won't actually help:

We can get an upper bound for the amount of answers by grouping the digits into sets e.g. $(5, 5)$ since the numbers must be in the same order, then replacing the numbers with $4^n$, to account for the 4 different operations we're allowed. You would have to figure out all the unique sets and sum up their totals. This would produce a huge amount of overlap however. $(8, 2)$ and $(8, 1, 1)$ will produce duplicates for many of it's operations. I can't think of a good way to trim them off the top of my head.


There are $9$ places to put the $4$ operations. So without brackets we have $4^9$ different representations. Though, some of the representations will be equal. For instance, $x\times1 = x\div1$.

The number of ways of inserting brackets into $n$ objects is given by the $n−1$ Catalan number. https://en.wikipedia.org/wiki/Catalan_number.

Having no brackets is the same as having brackets that induce the standard order of operations. The ninth Catalan number is $4862$. Therefore there are $4862\times4^9 = 1274544128$ distinct representations. However, if we remove the cases with $\div1$ at the end since this is the same as $\times1$. Then we have $4862\times3\times4^8 = 955\ 908\ 096$ representations. I'm not sure if any of these are equal.

  • $\begingroup$ Though more of them must be equal. For instance (3 + 2 + 1) = (3*2*1) $\endgroup$
    – gyancey
    Dec 31, 2019 at 20:40

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