Answer:
6
Step-by-step explanation:
If the question were how many distinct four digit number contain the digits 1,2,3, and 4 without repetition we could use the variation without repetition formula for 4 elements.
The first digit must be any of this four digits, so, we have 4 possibilities.
For the second digit, we already used one digit, so, we only have 3 possibilities. The total possibilities of picking this first two numbers are 4 * 3.
For the third digit, we already used two digits, so, we have 2 possibilities. The total possibilities thus far are then 4 * 3 * 2.
And for the final digit we don't have a choice, its the only one left. So we have 4 * 3 * 2 * 1 possibilities for a four digit number formed by the digits 1,2,3, and 4 without repetition .
This number its the factorial of 4:
[tex]4! \ = 4 \ * \ 3 \ * \ 2 \ * \ 1[/tex].
If the digit 2 is always immediately followed by the digit 3, no, we don't have 4 choices, cause, our elements are:
the digit 1 , the duple 2 3 , and the digit 4
So, for the first position, we must choose one of this elements. Lets say 1.
1 _ _ _.
Then, we can choose the duple 23 for the second and third position:
1 2 3 _
Finally we only got one choice, the digit 4:
1 2 3 4
So, as we can see. This is the same that choosing three elements without repetition. The possibilities will be:
[tex]3 \ * \ 2 \ * \ 1 \ = \ 3! \ = \ 6[/tex].
Now, we could: ask which are this possibilities?. They are:
1 2 3 4
1 4 2 3
2 3 1 4
2 3 4 1
4 1 2 3
4 2 3 1
