QJEEM23Mabc4b: JEE Main 2023 Mathematics - Class 11 JEEM23Mabc4b Solution: The total number of six digit numbers, formed using the digits…

A number will be divisible by 6 iff the digit at the unit place of the number is divisible by 2 and sum of all digits of the number is divisible by 3 .

Units, place must be occupied by 4 and hence, at least one 4 must be there.

Possible combination of 4, 5, 9 are as follows :

4	5	9	Total number of Number
1	1	4	${{5!} \over {4!}} = 5$
1	4	1	${{5!} \over {4!}} = 5$
2	2	2	${{5!} \over {2!2!}} = 30$
3	0	3	${{5!} \over {2!3!}} = 10$
3	3	0	${{5!} \over {2!3!}} = 10$
4	1	1	${{5!} \over {3!}} = 20$
6	0	0	${{5!} \over {5!}} = 1$

Total = 5 + 5 + 30 + 10 + 10 + 20 + 1 = 81.