Activity-permutations and combinations problems

Problem 1

How many 3-digits numbers can be formed from the digits 0,1,2,3 and 4 without repetation.(En.md.page no 69)

Interpretation:
3-digit number is to be formed using the numbers 0,1,2,3 and 4
Concepts:

1. when forming a 3-digit number '0' can't occupy the hundred place if it occupies then it becomes 2-digit number
2. how to fill the places wheater to start from the unit place or hundreds place (in case of the given numbers contain '0')
3. if one place is filled with an number then how many digits are left to be filled and how many numbers are there to be left for filling
4. application of fundamental principle of counting

Solution:
The digits in the selection set are {0,1,2,3,4}
A 3-digits number number will have three places i.e unit,tens,hundreds
If the selection set contains '0' then the filling must be started from the highest place, so in this problem we have to fill the hundreds place first.
Hundreds: from the given number of set hundred place can't be filled with '0' because it becomes a 2-digit number. Hence the hundreds place can be filled in by 1,2,3 or by 4 i.e in 4-ways

Tens: After filling a place we left with '0' and other 3 numbers (out of 1,2,3 and 4) so the tens place can be filled in 4 different way

Unit: Now after filling the 2 places the unit place can be filled in 3 different ways

Therefore by counting principle total number of ways that a 3-digit number can be formed using 0,1,2,3 and 4 are 4X4X3=48 numbers