How many ways can you add the numbers 1-5, USING ONLY 5 NUMBERS, to get a total of 15 or less. You may use numbers more than once

Question

1,1,1,1,1 is a possibility since it is less than 15. list the possible outcomes

1,2,1,1,1 is a different outcome than 1,1,2,1,1.

ORDER DOES MATTER

Answer 1

I cheated a bit too, and wrote a C program.  For each combination less than or equal to 15, I sorted the 5 numbers high to low and kept count of how many times that sorted value was seen (so for example  2 1 1 1 1    in the file means one 2 with four 1's where the 2 can be anywhere in the numbers picked).   A sequence like 2 1 1 1 1  is equivalent, as far as number of times it shows up, as "one number with four of some other number" if that makes any sense.

 

https://www.wyzant.com/resources/files/499876/adding_1_5_w_repeats_allowed

 

 

For the sequence 1 1 1 1 1  there is only one way to produce it (and since there is a constraint of sum <= 15, the only other two outputs where all five numbers are the same is 2 2 2 2 2 and 3 3 3 3 3).   [ this accounts for 3 arrangements ]

 

Outputs like 2 1 1 1 1   occur  5!/4! = 5 ways (5! ways to arrange 5 numbers, but we've over counted by 4! ways because the 1's are indistinguishable).    [ this pattern occurs 10 times ]

 

Then there are these additional five forms and their corresponding number of ways to produce them:

 2 2 1 1 1                    5!/(2!*3!) = 120/(2*6) = 10  ways   [ occurs 10 times ]

 5 2 1 1 1                    5!/(3!) = 120/6 = 20 ways               [ occurs 17 times ]

 3 3 2 2 1                    5!/(2!*2!) = 120/4 = 30 ways          [ occurs 18 times ]

 5 3 2 1 1                    5!/(2!) = 120/2 = 60 ways               [ occurs 10 times ]

 5 4 3 2 1                    5! = 120 ways                                 [ occurs 1 time ]

 

Where I get stuck is computing how many times we'd expect, say, two unique numbers with a number repeated 3 times, because it depends on the constraint of not adding to more than 15.   That form happens to appear 17 times.   Perhaps the setup requires more or less a brute-force approach...

 

 

Hope this helps.   

 

Answer 2

I cheated a bit and did this in a combo of R and Excel.  The answer is 1753 out of 3125 possible combinations.Unfortunately, I don't know how to mathematically prove that, but it is the answer nonetheless.