View source for Activity-permutations and combinations problems


= Problem 1 =
<br>
'''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:'''<br>
3-digit number is to be formed using the numbers 0,1,2,3 and 4<br>
'''Concepts:'''<br>
#when forming a 3-digit number '0' can't occupy the hundred place if it occupies then it becomes 2-digit number
#how to fill the places wheater to start from the unit place or hundreds place (in case of the given numbers contain '0')
#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
#application of fundamental principle of counting
'''Solution:'''<br>
The digits in the selection set are {0,1,2,3,4}<br>
A  3-digits number number will have three places i.e unit,tens,hundreds<br>
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.<br>
'''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<br>

'''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<br>

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

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


===formulas===
<math>\{lsup{8}}C_{2}</math>