# How Many Different Differences Can Be Obtained By Taking Only Two Numbers At A Time From 3,5,2,10 And 15?

Actually, the correct mathematical answer is 1,440, but you are on the right track by calculating the different differences first.

We see that:
The 1st #, three, has 4 different differences (choices).
The 2nd #, five, has 3 different differences (choices).
The 3rd #, two, has 4 different differences (choices).
The 4th #, ten, has 2 different differences (choices).
The 5th #, fifteen, has 3 different differences (choices).

However, we do not add together these differences to get 18.

Instead, we multiply: 4 x 3 x 4 x 2 x 3 = 288 total choices.
My explanation for this is there is no order as to which pair we subtract first. For example, we can begin with 3 - 5 or we can begin with 3 - 10. In this sense, order does not matter.

Lastly, we multiply: 288 x 5 = 1,440 total ways b/c again order does not matter. There are 5 different "sets" of combinations altogether. For example, we can  begin with 5 - 3 followed by 10 - 2.
thanked the writer.
Anonymous commented
how do u calculate the differences? I mean why
The 3rd #, two, has 4 different differences (choices).
The 4th #, ten, has 2 different differences (choices).
The 5th #, fifteen, has 3 different differences (choices).?
Any of the 5 numbers can be the first number, and any of the remaining 4 can be the second number, for a total of 20 differences. Because 10-5 = 15-10, there won't be that many different differences. Let d(x,y) represent the possible differences between x and y.
d(2,3) = ±1
d(2,5) = ±3
d(2,10) = ±8
d(2,15) = ±13
d(3,5) = ±2
d(3,10) = ±7
d(3,15) = ±12
d(5,10) = ±5
d(5,15) = ±10
d(10,15) = ±5

The different differences seem to be ±1, ±2, ±3, ±5, ±7, ±8, ±10, ±12, ±13, for a total of 18 different differences.
thanked the writer.
Rania Joseph commented
Guys how did u get the 18? 18 different differences?
Oddman commented
The answer lists them, so they are easy to count. +1, -1, +2, -2, +3, ...
Anonymous commented
i think u missed 0 