How many 4-digit odd numbers can be formed using the digits 0, 1, 2 and 3 only if the repetition of the digit is not allowed? As stated in the title above: How many 4-digit odd numbers can be formed using the digits 0, 1, 2 and 3 only if the repetition of the digit is not allowed?
I already have the answer for this and it is 8.
(2)(2)(1)(2) = 8
However, I do not understand how it became like that. Any understandable explanation would be appreciated!
 A: Hint #1: Since the number is odd, then the unit's digit could only be either $1$ or $3$.
Hint #2: Since the number has $4$ digits, the thousand's digit cannot be $0$.
Can you finish?
A: Here's how I would do it.  There are $4!=24$ permutations of the $4$ digits.  Half of them end in $0$ or $2$, so only $12$ of them are odd. Since a four-digit number can't start with $0$ we must exclude the permutations that start with $0$.
Once we fix the last digit, there are three digits left, and one of them is $0$, so we exclude one third of the $12$ cases.  That leaves $12-4=8$.
A: *

*Place the $0$ digit, but not first (no leading $0$) or last (not odd), with $2$ possibilities

*Choose an odd digit for the last place, with $2$ possibilities

*Place the $2$ digit in a remaining place, with $2$ possibilities

*Place the other odd digit in a remaining place, with $1$ possibility

*Multiply

A: Alternative approach.
Given any set of $k$ digits, including $0$ that are not to be repeated, the enumeration of how many $k$ digit numbers that can be formed is
$$k! - (k-1)!.\tag1$$
The second term expresses the elimination of $0$ as the leftmost digit.
In the present problem, the rightmost digit can be either $1$ or $3$.
Assume, without loss of generality, that it is $3$, complete the enumeration under that assumption, and then multiply the result by $2$, to reflect that the rightmost digit might also be $1$.
With $3$ as the rightmost digit, you then have $(4-1)$ digits remaining, including $0$.  Then employ equation (1) above, with $k=3$.
Therefore, the final enumeration is
$$2 \times [(3!) - (2!)].$$
A: The reason why the answer is (2)(2)(1)(2) is because

*

*First digit: Cannot be 0, so there are three remaining numbers 1,2,3. But you must reserve one number (1 or 3) for the last digit, so there are actually two possible numbers to choose for the first digit.


*Second digit: Once you have chosen a number for the first digit, you can now choose the number 0 and some different number than first digit also keeping in mind that another number (1 or 3) is reserved for the last digit. So there are two possible numbers to choose for the second digit.


*Third digit: Only one possible number for the third digit.


*Fourth digit: Two possible numbers for the fourth digit (1 or 3).
Using the product rule gives (2)(2)(1)(2).
An easier method is $4!-2(3!)-4=8$ where

*

*$4!$ is the total number of ways to arrange $0123$

*$2(3!)$ is the total number of ways to have the last digit of the number as $0$ or $2$

*$3!-2=4$ is the total number of ways to have the first digit of the number as $0$ excluding two cases where the number contains $2$ as the last digit

