How many six digit numbers can be formed using the digits 1 2 3 4 5 and 6 at most once such that the numbers are divisible by 3?

All such numbers with units digit 1,2,3,5 or 6 are divisible by that digit. 5! =120 ways each.

How many 6-digit number can be formed from the digit 1 2 3 4 5 6 which are divisible by 4 and digits are not repeated?

Therefore, there are a total of 192 numbers which can be formed using the digits 1,2,3,4,5,6 without repetition such that the number is divisible by 4. Hence, option (B) is the correct option.

How many 6 digits numbers can be formed with the digits 1 2 3 4 5 6 7 If the 10th unit’s places are always even and repetition is not allowed?

And possible digits left for fifth place will be 2. So, total number of numbers that can be formed with the digits {1, 2, 3, 4, 5, 6, 7} with no digits repeated and terminal digits as even will be 3×5×4×3×2×2 = 720. So, total possible numbers will be 720. Hence, the correct option will be D.

How many six digit number can be formed using the digits 1 to 6 which are divisible by 4?

∴ Total number of arrangements from both these cases is 48 + 48 = 96. Thus, 96 6-digit numbers can be formed that are divisible by 4.

How many 6-digit numbers can be formed using 6 digits?

∴ 360 different 6-digit numbers can be formed.

How many 6-digit even numbers can be formed from digits 1 to 7?

= 720 ways. Therefore, 720, six digit even numbers can be formed.

How many 6 digit numbers can be formed using digits 659942?

360 different 6 digit numbers can be formed using digits in the number 659942 , 96 of them are divisible by 4. = 6!/2!

How many 6 digit even numbers can be formed from digits 1 to 7?

How many 6 digit numbers can be formed from 1 to 7 so that the digits should not repeat and the second last digit is even?

Therefore, 720, six digit even numbers can be formed.

How many 6 digits can be made using 2 2’s and 2 3s?

We can select the positions of the two 2’s from the four remaining positions in (42) ways. We can select the position of the two 3’s from the two remaining positions in (22) ways. Thus, the number of six digit numbers with exactly two 1’s, two 2’s, and two 3’s is (62)(42)(22)=6! 4!