JEE Main 2023 — Permutation Combination Question with Solution

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, $\left\{\left(1,24\right),\left(3,22\right),....,\left(23,2\right)\right\}$

Total ordered pairs$=12$

If $\lambda =2$, then

$a+b=48$

No ordered pair possible.

If $\lambda =3$, then

$a+b=71$

So,

$\left\{\left(1,70\right), \left(3,68\right), \left(5,66\right),....,\left(61,10\right), \left(63,8\right), \left(65,6\right), \left(67,4\right), \left(69,2\right)\right\}$

Total $35$ ordered pairs.

If $\lambda =5$, then

$a+b=117$

$\left\{\left(17,100\right), \left(19,98\right), ..., \left(99,18\right)\right\}$

Total $42$ ordered pairs.

If $\lambda =7$, then

$a+b=163$

So,

$\left\{\left(63,100\right), \left(65,63\right),.... \left(99,64\right)\right\}$

Total $19$ pairs.

No further case possible.

So, required number is

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