How many numbers can be formed from the digits 1,2,3,4 when the repetition is not allowed
A. $^{4}{{\text{P}}_{4}}$
B. ${ }^{4} \mathrm{P}_{3}$
C. $^{4}{{\text{P}}_{1}}{{+}^{4}}{{\text{P}}_{2}}{{+}^{4}}{{\text{P}}_{3}}$
D. ${ }^{4} \mathrm{P}_{1}+{ }^{4} \mathrm{P}_{2}+{ }^{4} \mathrm{P}_{3}+{ }^{4} \mathrm{P}_{4}$

Hint: Basically A permutation is a grouping of items in a specific direction or sequence. One should consider both the selection and the arrangement while dealing with permutation. In permutations, ordering is crucially important. The permutation is seen as an ordered combination, in other words.

Complete step by step solution:
A permutation is a configuration of objects in a particular order. Here, the components of sets are arranged in a linear or sequential order. For instance, the set $A=1,6$ has a permutation of 2, which is 1,6,1. There is no other way to organize the components of set A, as you can see.
While the order of the ingredients isn't important in combination, it should be followed while performing permutations. Additionally, see Permutation and Combination.
Here in the given question,
There are 4 numbers . Then without repetition it can form numbers $^{4}{P}_{4}$ times
Consider $n=4$ and $r=4$ in order to analyze the aforementioned. and apply the permutations formula $p(n, r)={ }^{n}P_{r}=\dfrac{n !}{(n-r) !}$to determine how many combinations may be created by selecting 4 items from the total, which is likewise 4. All conceivable options will be covered in the response you receive.
$P(n, r)={ }^{n}P_{r}=\dfrac{n !}{(n-r) !}$
${ }^{4}P_{4}=\dfrac{4 \times 3 \times 2 \times 1}{1}$
${ }^{4}P_{4}=24$

Option ‘A’ is correct

Note: The number of maximum digit numbers formed using the digits 1, 2, 3, 4 when repetition is not allowed is equivalent to the number of 4 Letter permutations of 4 distinct letters.