How many three digit numbers can be formed using the digits 01 2 3 4 if repetition of digits is not allow?

With five digits to chose from and no repetitions, there are five possibilities for the first digit, four for the second, and three for the third, so there are

5 x 4 x 3 = 60 ways in total, but if you assume you can't have 0 as the first digit of a 3 digit number, this cuts out one option for the first digit leaving

4 x 4 x 3 = 48 ways in total

a. 4 x 4 x 4 = 64. You have four choices for each digit. b. You aren't explaining what you mean. All three-digit numbers have a ones and a tens digit, so it's the same answer: 64. c. 4 x 3 x 2 = 24. You have four choices for the first digit, three for the second (you can't repeat the first digit), and two choices for the last digit (you can't repeat the first two digits).

for forming a 3 digit no. using 5 digits i.e. 0,1,2,3,4 we can say

hundreds place can be filled by 4 digits that is 1,2,3,4

tens place can be filled by 4 digits that is 0 and the rest of the three digits not used in the hundreds place

and the unit place can be filled by 3 digits

hence no.of 3 digits formed is 4x4x3=48

1st digit 4 choices as 0 is not there

2nd digit 4 choices ( all except 1st digit)

3rd digiut 3 choices(all except 1st 2)

so number of numbers = 4*4*3 = 48

P(5, 3) -- P(4, 2) = 5*4*3 -- 4*3 = 60 -- 12 = 48

48

0,1,2,3 & 4 are 5 numbers so its like these

{0,1,2,3,4,}

2 squared 5=32 minus 1 because the other number will not have the number 0

the answer is 31 numbers