Question

Find the number of ways to choose an ordered pair $(a,b)$ of numbers from the set $\{1,2,3,\dots ,10\}$ such that $|a-b|\le 5.$
(correct answer + 5, wrong answer 0)

Answer 1

$a,b\in \{1,2,3,\dots ,10\}$
Let $A_i=\left\{\right(a,b):|a-b|=i\text{for}i=0,1,2,\dots ,5\}$
$A_0=\left\{\right(i,i)\text{for}i=0,1,2,\dots ,10\}$
$n\left(A_0\right)=10$

$A_1=\left\{\right(i,i+1)\text{for}i=1,2,\dots ,9\}$
We can swap $a$ and $b$ to get two distinct ordered pairs.
$\therefore n\left(A_1\right)=2\times 9=18$

$A_2=\left\{\right(i,i+2)\text{for}i=1,2,\dots ,8\}$
$\therefore n\left(A_2\right)=2\times 8=16$

$A_3=\left\{\right(i,i+3)\text{for}i=1,2,\dots ,7\}$
$\therefore n\left(A_3\right)=2\times 7=14$

$A_4=\left\{\right(i,i+4)\text{for}i=1,2,\dots ,6\}$
$\therefore n\left(A_4\right)=2\times 6=12$

$A_5=\left\{\right(i,i+5)\text{for}i=1,2,\dots ,5\}$
$\therefore n\left(A_5\right)=2\times 5=10$

$\therefore$ Required number of pairs
$=n\left(\bigcup _{i=0}^5{A}_i\right)$
$=\sum _{i=0}^{5}n\left(A_i\right)$
$=10+18+16+14+12+10$
$=80$

Alternate :
Total ordered pairs without restriction $=10\times 10=100$
Let $b-a\ge 6$
Now, if $1\le a<b\le 10,$
then $1\le a<b-5\le 5$
So, we have to choose $a$ and $b-5$ with $1\le a<b-5\le 5$
This can be done in ${}^5{C}_2=10$ ways.
Similarly, for $a-b\ge 6,$ there are $10$ ways.
Required number $=100-10-10=80$