3 letters without replacement 4 letters A B C D how many ways can this be done if the order of the choices matters

Since the order of choice matters, we will permute the values.                       a bit more explanation for this:

If the order of choice did NOT matter, ABC and BCA will be counted as one since order of choice does NOT matter

Since order of choice does matter, ABC , BCA and CAB are all different possibilities for the arrangement of the same 3 letters

Since we have 3 slots:

___  ___ ___

Now, for the first slot. You can out either one if the 4 alphabets in the first slot since no slot has been used as of now

So:

_4_ ___ ___

**Keep in mind that the 4 is the possible number of values this slot can have**

Now that one slot has been used, one of the 4 alphabets has been used and since we are not allowed to repeat the same alphabets, we are left with  3 more alphabets

we can put any one of the 3 alphabets in this second slot, Hence:

_4_ _3_ ___

Now that 2 of the 4 alphabets have been used, we are left with only 2 alphabets, so there are only 2 possible alphabets for slot 3

Therefore:

_4_ _3_ _2_

Now that we know the possible alphabets for all 3 slots, we will multiply them with each other to get the total possible number of 3 - alphabet words we can make with 4 alphabets

Total possible words = 4 * 3 * 2

Total possible words = 24

We could've used the formula for Permutation as well

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents