Permutations and combinations - choosing an integer

Question

Question:

In how many ways can we choose 2 distinct integers from 1 to 100 such that the difference between them is at most 10?

Approach: I tried to fix a certain number, and then find the number of integers that would satisfy the condition However, this is a very long method. Any help?

turkeyhundt · Accepted Answer · 2015-01-04T07:46:02.373

3

I would say for every number from 1 to 90, you can count the ten integers that follow it. $(1,2),(1,3),(1,4),\dot(1,11),(2,3),(2,4),\dots(2,12)\dots(90,100)$. Oh, then add in the 9 pairs starting with 91, the 8 pairs starting with 92, 7 starting with 93 etc... Ending with $(99,100)$ which I think is a total of 945.

$90\times10+\frac{9\times10}{2}$

I don't think that double counts anything or misses any, does it? I'll think it over for a second...

edited Jan 04 '15 at 07:46

answered Jan 04 '15 at 07:32

turkeyhundt

7,733

I came up with 945 too. But is there any easier method? – Gummy bears Jan 04 '15 at 07:42
There might be. Still turning it over... – turkeyhundt Jan 04 '15 at 07:46
1

The only other thing I can see is just taking every number from 1 to 99 and counting the 10 integers above them, $99\times10$ and then subtracting out the triangle of numbers that are greater than 100 $(45)$ to get $990-45=945$ – turkeyhundt Jan 04 '15 at 07:49

Permutations and combinations - choosing an integer

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