How many different pairs $(m, n)$ can be formed using numbers from the list of integers $\{1,2,3, \ldots, 20\}$ such that $m<n$ and $m+n$ is even?
(A) 55
(B) 90
(C) 140
(D) 110
(E) 50
The sum of two even numbers is even. The sum of two odd numbers is even.

The sum of an odd number and an even number is odd.

Thus, for the sum $m+n$ to be even, both $m$ and $n$ must be even, or they must both be odd.

If $m=2$, then $n$ must be even and greater than 2 .

Thus, if $m=2$ then $n$ can be $4,6,8,10,12,14,16,18$, or 20 .

This gives 9 different pairs $(m, n)$ when $m=2$.

If $m=4$, then $n$ must be even and greater than 4 .

Thus, if $m=4$ then $n$ can be $6,8,10,12,14,16,18$, or 20 .

This gives 8 different pairs $(m, n)$ when $m=4$.

Continuing in this manner, each time we increase $m$ by 2 , the number of choices for $n$, and thus for $(m, n)$, decreases by 1 .

This continues until $m=18$, at which point there is only one choice for $n$, namely $n=20$.

Therefore, the total number of different pairs $(m, n)$ where both $m$ and $n$ are even is,

$9+8+7+6+5+4+3+2+1=45$.

Similarly, if $m=1$, then $n$ must be odd and greater than 1 .

Thus, if $m=1$, then $n$ can be $3,5,7,9,11,13,15,17$, or 19 .

This gives 9 different pairs $(m, n)$ when $m=1$.

If $m=3$, then $n$ must be odd and greater than 3 .

Thus, if $m=3$ then $n$ can be $5,7,9,11,13,15,17$, or 19 .

This gives 8 different pairs $(m, n)$ when $m=3$.

Continuing in this manner, each time we increase $m$ by 2 , the number of choices for $n$, and thus for $(m, n)$, decreases by 1 .

This continues until $m=17$, at which point there is only one choice for $n$, namely $n=19$.

Therefore, the total number of different pairs $(m, n)$ where both $m$ and $n$ are odd is,

$9+8+7+6+5+4+3+2+1=45$.

Thus, the total number of different pairs $(m, n)$ using numbers from the list $\{1,2,3, \ldots, 20\}$ such that $m<n$ and $m+n$ is even is $45+45=90$.

ANSWER: (B)