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 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 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-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 numbers can you make with 6 digits?

There are 999999 – 100000 + 1 possible ways to make 6 digit numbers using the dgits 1 – 9. There are 900000 possible ways to get 6 digit numbers using 0 – 9.

How many 6 digit numbers can be formed from the digits 112233?

3 = 90 Answer.

How many 5 digit prime numbers can be formed without repetition?

Prime numbers are numbers which only divisible by 1 and itself. So, the number of 5 digit prime numbers formed using the digits 1, 2, 3, 4, and 5 using repetition is 0. Here we need to focus on those five digit numbers formed by all the digits 1, 2, 3, 4 and 5 without repetition.

How many 5 digit numbers are completely divisible by 3?

So, sum of the digits of every five digit number such formed = (1 + 2 + 3 + 4 + 5) = 15; which is a multiple of 3. Thus, all five digit numbers such formed must be completely divisible by 3 (Refer divisibility rule of 3).

What is the total number of possible 3 digit numbers?

Now, as the repetition is allowed, the digits possible at B is 6. Similarly, at A, also, the number of digits possible is 6. Therefore, The total number possible 3 digit numbers = 6 × 6 × 3 = 108. 3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

How many four digit numbers can be there with no repeated?

At ten’s place = There are now 8 digits to be filled as 2 digits are already taken = 8 ways Total Number of ways to fill the four places = 9 × 9 × 8 × 7 = 4536 ways. So a total of 4536 four digit numbers can be there with no digits repeated.