How many 6 digit numbers can be formed if no digit is repeated?

Similarly, if all the digits from 1 to 9 were allowed (with no digit repeated), then the answer would be 9*8*7*6*5*4 = 60480. The tricky part comes when the digit 0 is allowed: suppose that you wanted to form a 6-digit number using digits from 0,1,…,9 without repeating a digit.

How many 6 digit numbers without repetition of digits are there such that the digits are all non zero and 1 and 2 do not appear consecutively in either order?

b ) 1 and 2 do not appear consecutively in either order? But the answer is 52080.

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 digit even numbers can be formed from dhow many 6 digit even numbers can be formed from digits 1 2 3 4 5 SO digit The digit repeat not is even?

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

How many 6 digit even numbers can be formed from the digits 1 2 3 4 5 7 the digits should not repeat and second final digit is even select one?

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 4 digit numbers can be formed without repeating any digit?

How many 4 digit numbers can be formed without repeating any digit from the following digits: 1, 2, 3, 4 and 6? An error occurred. Please try again later There are 120 – 4 digit numbers can be formed without repeating any digit

How many possible digits are there for a 6-digit number?

A 6-digit number cannot have 0 in the leftmost space, therefore there are 5 possible digits for this space (1,2,3,4 or 5).

What is the first 2nd and 3rd digit of a number?

For numbers formed ending with 0, the 1st digit can be one of (1, 2, 3, 4, 5) = 5 choices. The 2nd digit can be one of the remaining 4 = 4 choices. The 3rd digit can be one of the remaining 3 = 3 choices.

How do you fill a 5 digit number with 9 numbers?

Lets say the five digit number is ‘abcde’, now ‘a’ can be any of the 9 numbers among (1,2,3…9), since 0 can’t be at the beginning. The second place, ‘b’ can br filled with any of the 9 numbers remaining. Similarly, ‘c’ can be filled with remaining 8 numbers, ‘d’ can be filled with 7 numbers and ‘e’ can be filled with 6 numbers.