How many 4 digits numbers not exceeding 4321 can be formed with the digits 1, 2, 3, 4 if the repetition of digits in a number is not allowed?

Hint: We will be using the concept of permutation and combination to solve the problem. We will use the multiplication rule of finding all possible combinations to further simplify the problem.

Complete step-by-step answer:
Now, we have to find 4 digit numbers which are not exceeding 4321 and can be formed with the digits 1, 2, 3, 4.
Now, if we represents the digit of the number then we have four boxes as,
\[\]
Since, the number has to be always less than that of 4321 and the digits as 1, 2, 3, 4 and also if repetition of digits in a number is not allowed.
So, if we keep 3 at thousands places then the total number starting with 3 is 3!.
Also, if we keep 2 at thousands places then the total number starting with 2 is 3!.
Similarly, if we keep 1 at thousands places then the total number starting with 1 is 3!.
Now, the numbers which are less than 4321 and starting with 42 and 41 are $2\times 2!$.
Also, there is one case when the number is starting with 43 and the number is less than 4321 is 4312.
So, total numbers which are less than 4321 are,
$\begin{align}
  & =3!+3!+3!+2\times 2!+1 \\
 & =6+6+6+4+1 \\
 & =18+5 \\
 & =23 \\
\end{align}$
Therefore, the answer is 23.

Note: To solve these types of questions it is important to note that the digits are not repeating. So, the permutation will be done by removing the digits which are already occupied at a place. Therefore, to find all the numbers starting with 3 we have done,
$3\times 2\times 1=3!$