- #1

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- 3

There are 5 letters, ABCDE. How many unique combinations of 3 letters can be made if letter A must be part of the combinations?

The solution is that because A must be one of the letters, that only leaves two other letters which you can manipulate. So the answer is (4)(3)/(2!) = 12/2 = 6 possible unique combinations

But my confusion arises from the explanation as to WHY the answer is calculated this way.

For example, if the condition that A must be part of the combination, then I know the three letter combination has 5*4*3/(3!) = 10 unique combinations.

But when I consider that A must be part of the solution, I think like this:

1) There are 3 boxes, and one of them must be A, meaning it can only be one possible choice.

2) The rest of the two boxes can be the other 4 choices.

3) So my permutations are 1 * 4 *3 = 12

4) Unique combinations are 12/(3!) = 2

**When I write out the solution, I know that my answer of 2 is wrong, and the answer is 6 is correct, but I cannot understand the reasoning behind calculating 6 the way it's done.**

I can slightly reconcile the reasoning if I think like this:

1) I need an A in the answer, and that can be made by choosing the option A only, which is 1 option, and the unique combination of that is 1/1! = 1

2) The rest of the two boxes can be the other 4 letters, so the unique combination of the last two boxes are 4*3/2!

3) The total unique combination is Ways to Choose A * Ways to Choose Other Two Letters = 1/(1!) * 4*3/(2!) = 6

But still, I'm a bit confused :(