QJEEM23Mac4cb: JEE Main 2023 Mathematics - Class 11 JEEM23Mac4cb Solution: The number of permutations, of the digits 1, 2, 3, ..., 7…

Given that digits are $1,2,3,4,5,6,7$

Total permutations $=7$ !

Let $p=$ Number which containing string 153

$q=$ Number which containing string 2467

$$\begin{array}{ll}& \therefore n(p)=5!\times 1\\ \\ & \Rightarrow n(q)=4!\times 1\\ \\ & \Rightarrow n(p\cap q)=2!\end{array}$$

$$\begin{array}{rl}& \therefore n(p\cup q)=n(p)+n(q)-n(p\cap q)\\ \\ & =5!+4!-2!=120+24-2=142\end{array}$$

$\therefore n$ (neither string 143 nor string 2467)

$$=7!-142=5040-142=4898$$