from the numbers 0,1,2,3,4 and 5 four digits number will be made, how many combinations can be made if the numbers aren't overlapped and odd (A) aren't overlapped and even (B)
im confused because in the even numbers, i used the box method and i got lots of answers.
my first answer is 156, i got it by making two boxes with for small boxes
on the first box : [5][4][3][1] and this is for numbers that end in zero
on the second box : [4][4][3][2] this is for numbers that end in 2 and 4
so total is 156
then i searched for the odd numbers, i got 144 with [4][4][3][3]
from what i've thought the odd and even numbers have same quantities, like if there are n numbers of odd number combination, there will be n numbers of even number combination
can someone help me solve this? i really appreciate your help. thanks
Best Answer
The reason for the difference in the quantity of odds and the quantity of evens is $0$. Instead of looking at four digits numbers, look at arbitrary strings of four digits that don't repeat digits. That means you can get $1234$, or $5320$, but also $0123$. Among these strings, there are as many that end with an odd number as there are strings that end with an even number. These are the same quantities you expected to see among the four digit numbers.
Now look at all of the strings of four digits that start with $0$, like $0243$ and $0512$. More of these strings end with an odd digit rather than an even digit, since $0$ has already been placed, so there are $3$ odd digits that it can end with, and only $2$ even digits that it can end with. You have to delete these strings to get the four digit numbers that you want to count. Since you're deleting more odd strings than even strings, you'll have more even strings at the end.