Help! Hard math problem.

Question

Kevin will start with the integers 1, 2, 3 and 4 each used exactly once and written in a row in any order. Then he will find the sum of the adjacent pairs of integers in each row to make a new row, until one integer is left. For example, if he starts with 3, 2, 1, 4, then he takes sums to get 5, 3, 5, followed by 8, 8, and he ends with the final sum 16. Including all of Kevin's possible starting arrangements of the integers 1, 2, 3 and 4, how many possible final sums are there?

Answer 1

First, let's figure out how many different ways Kevin could write those integers. He can write any of the four integers first, then any of the three remaining integers, then either of the two remaining integers, and then the last integer for a total of 4×3×2×1 = 24 ways to write the numbers.

 

Those 24 ways are:

 

1 2 3 4

1 2 4 3

1 3 2 4

1 3 4 2

1 4 2 3

1 4 3 2

 

2 1 3 4

2 1 4 3

2 3 1 4

2 3 4 1

2 4 1 3

2 4 3 1

 

3 1 2 4

3 1 4 2

3 2 1 4

3 2 4 1

3 4 1 2

3 4 2 1

 

4 1 2 3

4 1 3 2

4 2 1 3

4 2 3 1

4 3 1 2

4 3 2 1

 

 

Some of these will have the same sums, though. In particular, any two that are exactly flipped, like 1, 2, 3, 4 and 4, 3, 2, 1:

 

1 2 3 4

 3 5 7

  8 12

   20

 

4 3 2 1

 7 5 3

  12 8

   20

 

 

So, you can eliminate 12 of the ways, since they'll definitely have duplicate sums. You'll be left with:

 

1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2

2 1 3 4
2 1 4 3
2 3 1 4
2 4 1 3

3 1 2 4
3 2 1 4

 

 

You can calculate those 12 sums and figure out whether there are any duplicates left!