JEE 2022 — Mathematics PYQ

Explanation

$b_i\in \{1,2,3,\dots ,100\}$

Let $P$ = set when $b_1,b_2,b_3$ are consecutive.

$\therefore n(P)=\underset{98\text{ times}}{\underbrace{97+97+97+\cdots +97}}=97\times 98$

Let $Q$ = set when $b_2,b_3,b_4$ are consecutive.

$\therefore n(Q)=\underset{98\text{ times}}{\underbrace{97+97+\cdots +97}}=97\times 98$

Now, $P\cap Q$ = set when $b_1,b_2,b_3,b_4$ are consecutive.

So, $n(P\cup Q)=n(P)+n(Q)-n(P\cap Q)$

$=97\times 98+97\times 98-97$

$=97(98+98-1)$

$=97(195)$

$=18915$