How many ways are there to select an ordered pair of numbers from $1$ to $7$ so that the sum is even? How many ways are there to select an ordered pair of numbers from $1$ to $7$ so that the sum is even?
Soln: The way I tried approaching this problem was, I made a grid of $7 \times 7$ squares. So $49$ ordered pairs in total.  Also the sum of two numbers is even only if the numbers selected are either both even or both odd. So, how do I proceed further?
Thank you.
 A: If we are allowed to select each element once, then add up the following:


*

*The number of ways to select an ordered pair from $\{2,4,6\}$, which is $3\cdot2=6$

*The number of ways to select an ordered pair from $\{1,3,5,7\}$, which is $4\cdot3=12$



If we are allowed to select each element twice, then add up the following:


*

*The number of ways to select an ordered pair from $\{2,4,6\}$, which is $3\cdot3=9$

*The number of ways to select an ordered pair from $\{1,3,5,7\}$, which is $4\cdot4=16$

A: Your table is
$\begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|}  \hline \text{sum }  & 1 &2 &3 &4 &5 &6 & 7\\ \hline \hline \hline 1 &\color{red}2 &3 &\color{red}4 &5 &\color{red}6 &7&\color{red}8 \\  \hline 2 & 3 &\color{red}4 &5 &\color{red}6 &7&\color{red}8 &9 \\ \hline 3&\color{red}4 &5 &\color{red}6 &7 &\color{red}8&9&\color{red}{10} \\ \hline 4 &5 &\color{red}6 &7&\color{red}8&9&\color{red}{10}&11  \\ \hline  5 &\color{red}6 &7&\color{red}8&9&\color{red}{10}&11&\color{red}{12}  \\ \hline 6 &7&\color{red}8&9&\color{red}{10}&11&\color{red}{12}&13 \\ \hline  7&\color{red}8&9&\color{red}{10}&11&\color{red}{12}&13&\color{red}{14}  \\ \hline \end{array}$
The red marked numbers are the even sums. Now you can sum up the up-right diagonals of the red marked numbers: $1+3+5+7+5+3+1=25$
Additional you can calculate the probability of selecting an even sum.
The number of possible outcomes is $7\cdot 7=49$
Therefore the probability to select an ordered pair of numbers from 
1 to 7 so that the sum is even is $\frac{25}{49}\approx 51\%$
