What is the number of 3 digit number that can be formed using the digits 0 1and 5 without repeating a digit in any number?

Question 8 How many different 3-digit numbers can be formed by using the digits 0, 2, 5 without repeating any digit in the number? The different 3-digit numbers which can be formed by using the digits 0, 2, 5 without repeating any digit in the number are 205, 250, 502 and 520.

How many 3 digit numbers are there including 0?

In the tens place and the one’s place, there can be any digit from 0-9. There is a possibility of 10 digits in the hundredth place and 10 digits in the unit’s place. Thus, there are 9 × 10 × 10 = 900 three-digit numbers in all. Therefore there are 900 three-digit numbers in all.

How many 3 digit numbers can be formed by using the digits 1 to 9 if digits can be repeated?

Hence, the required number of numbers = 504.

What is the sum of all four digit numbers formed?

Now, sum of all four digit numbers formed = Sum of all numbers − Sum of numbers with 0 in the first place. Hence, the answer = 66660−2220= 64440. Was this answer helpful?

What is the sum of total numbers ending with 2?

Total numbers ending with 2 is 3! as after fixing 2 in the unit’s place other three places can be filled by 3! ways. Thus, 2 appears in the unit’s place 3! times. Similarly, all other digits 4,6 and 8 also appear 3! times. Then sum of the digits in the unit’s place is 6(2+4+6+8)=120 units .

How many five-digit numbers are there using the given digits?

If 5 is at the unit’s place, the remaining digits 1,2,3 and 4 can be arranged among themselves in 4! ways. Hence, there are exactly 4! five-digit numbers using the given digits. Similarly, there are exactly 4! five-digit numbers using the given digits.

What is the sum of all the 24 numbers?

Similarly, sum of digits in ten’s place is 120 tens and that in hundred’s place is 120 hundreds, etc. Hence, sum of all the 24 numbers is 120(1+10+10 2+10 3)=120×1111=133320. Was this answer helpful?