There are 10 players to choose from and you need to choose 5.
The difference between part A and part B is that in A positions are NOT assigned so every person, regardless of when they are chosen has the same role. In part B they have different roles (positions).
For a second let's present there are 2 people (Jay and Jo) and we need to choose 2 of them. If we don't assign them a position there is only one way to do this, you have to take both people Jay and Jo. Let's assume now that there are positions (offense and defense)...now there are 2 ways to pick them...Jay for offense and Jo for defense or Jay for defense and Jo for offense. The order in which we pick them matters in the second scenario because each pick means a particular position. We also get more options this way.
Mathematically in A you have a situation that can be modeled with a combination and in B with a permutation.
PART A
Use the combination nCr as order does not matter. This is a fraction. [tex]10C5= \frac{10!}{5!(10-5)!}[/tex]
In symbols you have: [tex]nCr= \frac{n!}{r!(n-r)!} [/tex].
Recall, n! = (n)(n-1)(n-2)(n-3)...(1)
In this case you get: [tex]10C5= \frac{10!}{5!(10-5)!}[/tex]
[tex]= \frac{10!}{5!5!}[/tex]
[tex] \frac{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)}{(5)(4)(3)(2)(1)(5)(4)(3)(2)(1)} [/tex]
Any number appearing in the numerator and denominator can be crossed off (divided out) so you end up with:
[tex] \frac{(10)(9)(8)(7)(6)}{(5)(4)(3)(2)(1)} = 252[/tex]
PART B
Here the order matters so we use a permutation. Specifically, nPr which we can evaluate this way (start with n and count down r terms). Here we obtain: [tex]10P5=(10)(9)(8)(7)(6)=151,200[/tex]
