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How many ways can a pair of integers be selected from the integers 1 to 99 so that the difference between them is 10?
If the order of the pairs of numbers doesn't matter (that is the pairs (1, 11) and (11, 1) are considered the same) then there are 89 such pairs.
If the order does matter (that is the pairs (1, 11) and (11, 1) are considered different) then there are 178 such pairs.
There must be one number larger than the other.
The smaller number must be at least 1, so the larger number must be at least 11.
The larger number must be at most 99, so the smaller number must be at most 89.
Thus, looking at the smaller numbers, it can be one of the numbers 1 to 89 - a total of 89 possibilities - each of which has a corresponding larger number.
Therefore there are 89 combinations of 2 integers from the number 1-99 that have a difference of 10.
If however, the order of the integers is important, then the smaller number can be the first number or second number, which means that there are twice as many possible pairs, giving 178 permutations of 2 integers from the numbers 1-99 that have a difference of 10.
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