The number of ways of selecting two numbers a and b , a ∈ \ 2, 4,...

Let $a = \left\{2, 4, 6, . . . , 100\right\}$ and $b = \left\{1, 3, 5, . . . , 99\right\}$

Let

$a + b = 23\lambda + 2$

If $\lambda = 1$, then

$a + b = 25$

So, \left\{\left(1, 24\right), \left(3, 22\right), . . . . , \left(23, 2\right)\right\}$</p><p>Total ordered pairs$= 12$</p><p>If$\lambda = 2$, then</p><p>$a + b = 48$</p><p>No ordered pair possible.</p><p>If$\lambda = 3$, then</p><p>$a + b = 71$</p><p>So,</p><p>\left\{\left(1, 70\right), \left(3, 68\right), \left(5, 66\right), . . . . , \left(61, 10\right), \left(63, 8\right), \left(65, 6\right), \left(67, 4\right), \left(69, 2\right)\right\}$

Total $35$ ordered pairs.

If $\lambda = 5$, then

$a + b = 117$

\left\{\left(17, 100\right), \left(19, 98\right), . . . , \left(99, 18\right)\right\}$</p><p>Total$42$ordered pairs.</p><p>If$\lambda = 7$, then</p><p>$a + b = 163$</p><p>So,</p><p>\left\{\left(63, 100\right), \left(65, 63\right), . . . . \left(99, 64\right)\right\}$

Total $19$ pairs.

No further case possible.

So, required number is

$= 12 + 35 + 42 + 19 = 108$