Combantrics Anagram Set Up

[ghong14]

I have been using the anagram set up to solve the comb. problems and I needed some clarification in terms of the method.

How many ways can be pick 3 people to play in a basketball team out of five people?

12345
YYYNN

So the calculation would be 5!/(3!2!)

On the other hand in the following problem that set up would be different.

How many ways can be pick a Front guard, middle guard and Back guard out of five players?

12345
FMBNN

So the calculation would be 5!/2!

However what I don't understand is how do we know that when something is distinct or not. For example have the specific position on the later is obviously specific therefore we did not put FMB as merely YYY (Y for being picked). However what about this.

How many way can you pick 2 people out of 5?

Is it

12345
YYNNN

5!/(2!3!)

or

12345
FSNNN

5!/3!

How do I differentiate them in the future?

[tim]

your final example gives two scenarios, one where we pick two people to be a group of two people, the other where we pick a person to be F (presumably first) and the second to be S. you brought in a distinction that may or may not be relevant depending on the context. in general, you should ask yourself if the selected items will be in distinct (i.e. distinguishable) groups. if you're selecting them to be part of the same group (i.e. you can't tell apart their role) then give them the same letter..