How many distinct four digit numbers contain the digits 1,2,3, and 4 without repetition, if the digit 2 is always immediately followed by the digit 3?

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