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Number of ways of choosing 2 distinct natural numbers $\leq 100$, such that they differ by at most $10$

My Way

Let the numbers be $x$ and $y$

               X                   Y
 <------a------><---------b--------><---------c------->

$a+b+c=100-2$; $a,c \geq 0$; $0 \leq b \leq 9$

Please Help.

  • 1
    integers or reals? – 2017-01-29
  • 0
    Sorry my mistake x and y are natural numbers – 2017-01-29

3 Answers 3

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Assuming you mean naturals, suppose $y>x$. For $1\leq x \leq 90$ there are 10 possible values of $y$ for each $x$. For $x = 91$ there are $9$ values of $y$, $x=92$, $8$ values, and so on. Thus if $y>x$ there are $$90\times 10 + 9 + 8 + ... + 1 = 945$$ ways to pick your two numbers. To include the case $y

EDIT: Sorry, doesn't look like you were looking for an ordered pair, so no need to multiply by two!

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There are different ways to tackle this question, but let's try your way:

If $b=0$, how many ways can $a$ and $c$ be chosen?

If $b=1$, how many ways can $a$ and $c$ be chosen?

Continue onward up to $b=9$. Then add up all the ways.

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For difference at least 11, we could calculate: Pick two different numbers in $\{1,2,\ldots,90\}$ and add $10$ to the larger one. That's possible in $90\choose 2$ ways. If we substract this from the $100\choose 2$ ways to pick two differentt numbers from $\{1,\ldots,100\}$, we have the desired result.