QJEEM23Mac3b8: JEE Main 2023 Mathematics - Class 11 JEEM23Mac3b8 Solution: The number of 9 digit numbers, that can be formed using all the …

Here, even digits are 2 and 4.

Number of digit "2" presents = 2

Number of digit "4" presents = 2

$\therefore$ Total even digits = 4

$\therefore$ Total 4 even places presents.

Number of ways to put those 4 digits in those 4 places $= {{{}^4{C_4} \times 4!} \over {2!2!}} = {{4!} \over {2!2!}}$

Now, remaining 5 digits (three 1 and two 3) can be put in those 5 places in $= {{{}^5{C_5} \times 5!} \over {3!2!}} = {{5!} \over {3!2!}}$ ways.

$\therefore$ Total possible 9 digit numbers

$= {{4!} \over {2!2!}} \times {{5!} \over {3!2!}} = 60$