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

Answer:

24 final sums, different sums are 16, 18, 20, 22 or 24.

Step-by-step explanation:

First, count how many different orders of 1, 2, 3 and 4 are possible. The first number can be chosen in 4 different ways (it can be any of 4 numbers), the second - in 3 different ways (it can be any of remaining 3 numbers), the third - in 2 different ways (any of remaining 2 numbers) and the last number - in 1 way (only one number left). In total,

$4\cdot 3\cdot 2\cdot 1=24$

different orders.

For each order we can

• write 3 sums in the first step;
• write 2 sums in the second step;
• write 1 sum in the last step.

In total, 6 sums (only 1 of them is final). So, there are 24 final sums.

$1, 2, 3, 4\rightarrow 3,5,7\rightarrow 8,12\rightarrow 20\\ \\1, 2, 4, 3\rightarrow 3,6,7\rightarrow 9,13\rightarrow 22\\\\1, 3, 2, 4\rightarrow 4,5,6\rightarrow 9,11\rightarrow 20\\\\1, 3, 4, 2\rightarrow 4,7,6\rightarrow 11,13\rightarrow 24\\\\3,1,2,4\rightarrow 4,3,6\rightarrow 7,9\rightarrow 16\\ \\3, 2, 1, 4\rightarrow 5,3,5\rightarrow 8,8\rightarrow 16\\\\3, 2, 4, 1\rightarrow 5,6,5\rightarrow 11,11\rightarrow 22\\\\3, 4, 1, 2\rightarrow 7,5,3\rightarrow 12,8\rightarrow 20$

$3, 4, 2, 1\rightarrow 7,6,3\rightarrow 13,9\rightarrow 22\\ \\3, 1, 4, 2\rightarrow 4,5,6\rightarrow 9,11\rightarrow 20\\\\1, 4, 2, 3\rightarrow 5,6,5\rightarrow 11,11\rightarrow 22\\\\1, 4, 3, 2\rightarrow 5,7,5\rightarrow 12,12\rightarrow 24\\\\4,1,2,3\rightarrow 5,3,5\rightarrow 8,8\rightarrow 16\\ \\4, 1, 3, 2\rightarrow 5,4,5\rightarrow 9,9\rightarrow 18\\\\4, 2, 3, 1\rightarrow 6,5,4\rightarrow 11,9\rightarrow 20\\\\4, 2, 1, 3\rightarrow 6,3,4\rightarrow 9,7\rightarrow 16$

$4, 3, 2, 1\rightarrow 7,5,3\rightarrow 12,8\rightarrow 20\\ \\4, 3, 1, 2\rightarrow 7,4,3\rightarrow 11,7\rightarrow 18\\\\2, 1, 3, 4\rightarrow 3,4,7\rightarrow 7,11\rightarrow 18\\\\2, 1, 4, 3\rightarrow 3,5,7\rightarrow 8,12\rightarrow 20\\\\2,3,1,4\rightarrow 5,4,5\rightarrow 9,9\rightarrow 18\\ \\2, 3, 4, 1\rightarrow 5,7,5\rightarrow 12,12\rightarrow 24\\\\2, 4, 3, 1\rightarrow 6,7,4\rightarrow 13,11\rightarrow 24\\\\2, 4, 1, 3\rightarrow 6,5,4\rightarrow 11,9\rightarrow 20$

Different final sums are 16, 18, 20, 22 and 24.