Here’s another effort at tackling instruction around counting problems. Inspired by James Tanton’s great work over here!
You have 5 players on a basketball team. We shall refer to them by their jersey numbers: {1,2,3,4,5}.
 The team is travelling to an away game on a bus. The players will sit in a line, one in front of the other. How many seating arrangements of these 5 players are possible?
 The coach needs to assign positions to each of these players. 2 of them will be guards and 3 of them will be forwards. How many assignments are possible?
 The coach needs to pick a point guard and a shooting guard. How many assignments are possible?
There are perhaps several ways to approach these problems. We seek methods that satisfy these criteria:
 Generalization. We prefer a method that can be used to solve other, similar problems.
 Unification. We prefer a method that will unify the three scenarios, so that we can treat them all just as “counting problems!”
Scenario #1: Seating Arrangements
Seat A 
Seat B 
Seat C 
Seat D 
Seat E 
1 
2 
3 
4 
5 
1 
2 
3 
5 
4 
1 
2 
4 
3 
5 
1 
2 
4 
5 
3 
… 
… 
… 
… 
… 
Analysis:
 First we asssign a player to seat A – we have 5 choices.
 Next we assign a player to seat B – since one player is assigned to seat A, we have 4 choices remaining.
 In the same way, there are 3 choices for C, 2 for D, and only 1 left for E.
 Thus there are 5*4*3*2*1 seating arrangements, or 5! = 120 possibilities.
Scenario #2: Assigning Positions
 We’ll use “F” for forwards and “G” for guards. This is starting to feel like a labeling problem!

FFF.GG 
FFF.GG 
FFF.GG 
FFF.GG 
FFF.GG 
FFF.GG 
Row 1 
123.45 
132.45 
213.45 
231.45 
312.45 
321.45 
Row 2 
123.54 
132.54 
213.54 
231.54 
312.54 
321.54 
Row 3 
124.35 
142.35 
214.35 
241.35 
412.35 
421.35 
Row 4 
124.53 
142.53 
214.53 
241.53 
412.53 
421.53 
Row 5 
… 
… 
… 
… 
… 
… 
Row 6 
… 
… 
… 
… 
… 
… 
[Note: these tables were originally developed in Word, and the rows were colorcoded. If anyone wants the file, email me and I’ll send it to you. So in the text that follows, the “blueblock” is the first two rows, the “yellowblock” is the next two rows, etc.]
Analysis:
 If we simply continue to list every combination of players as we did in the first problem, we would see something like the table above.
 In that case we would again have a table with 120 cells, but scenario #2 certainly has fewer outcomes than this. What are we to do to correct this?
 Let’s examine the table very closely. In rows 1 and 2, players {1,2,3} are forwards, and {4,5} are guards. In row 1, the guards are in the same “seats” and the forwards are shuffled. Since we are shuffling 3 players around, there are 3*2*1, or 3! = 6 arrangements.
 In row 2, the forwards are in the same “seats” as row 1, but the guards have been shuffled. Since there were 2 players to shuffle, there are 2*1, or 2! = 2 arrangements.
 Summing up: in this scenario we are only interesed in the positions assigned to each player, so all of the blue arrangements are *equivalent,* all of the yellow are equivalent, and so on.
 There are (3!)(2!) cells in the blue block, and also the yellow block, and so on.
 Thus to count the number of playerassignments possible in this scenario, we take the number of cells in the whole table, which was 5!, and divide by the size of each colorblock, which is 3!2!. So we believe there ought to be 5! / (3! 2!)ways to choose 3 forwards and 2 guards from a group of 5 players. That gives 5*4 / 2 = 5*2 = 10 choices for the coach this time around!
Scenario #3: Choosing Guards
 We’ll use “P” for point guard and “S” for shooting guard. But what about the other players? We’ll just use “N” for “not a guard,” but they are probably forwards!

NNN.P.S. 
NNN.P.S. 
NNN.P.S. 
NNN.P.S. 
NNN.P.S. 
NNN.P.S. 
Row 1 
123.45 
132.45 
213.45 
231.45 
312.45 
321.45 
Row 2 
123.54 
132.54 
213.54 
231.54 
312.54 
321.54 
Row 3 
124.35 
142.35 
214.35 
241.35 
412.35 
421.35 
Row 4 
124.53 
142.53 
214.53 
241.53 
412.53 
421.53 
Row 5 
… 
… 
… 
… 
… 
… 
Row 6 
… 
… 
… 
… 
… 
… 
Analysis:
 The analysis runs similar to scenario #2, but this time we make a distinction between types of guards: are you a point guard or a shooting guard? What are the implications of this?
 First we note that the cells in the blue block are equivalent with respect to the “nonguards.” However, this time the light blue cells in row 1 are different from the dark blue cells in row 2, since assigning player #4 to be the PG and #5 to be the SG is different from putting #5 at PG and #4 at SG.
 We conclude that the cells in a given row are “the same,” but the rows are all distinct cases.
 Thus we have 5! / 3! = 5*4 = 20 ways to accomplish task #3.
 In fact, if we consider that there are “1 arrangements” in each column for the point guard, and “1 arrangements” for the shooting guard, then we have an expression in which every player is properly labeled: 5! / (3! 1! 1! 1!).
Synthesis
Let us now revisit the original problems, and try to solve all of them with a single strategy: they are simply labeling problems!
You have 5 players on a basketball team. We shall refer to them by their jersey numbers: {1,2,3,4,5}.
#1. The team is travelling to an away game on a bus. The players will sit in a line, one in front of the other. How many seating arrangements of these 5 players are possible?
Solution: There are 5! arrangements, and there are 5 labels – one for each seat: {A, B, C, D, E}. Thus there are 5! / (1! 1! 1! 1! 1!) = 120 outcomes.
#2. The coach needs to assign positions to each of these players. 2 of them will be guards and 3 of them will be forwards. How many assignments are possible?
Solution: There are 5! arrangements, and there are 2 labels – {G,G} and {F,F,F}. Thus there are 5! / (2! 3!) = 10 outcomes.
#3. The coach needs to pick a point guard and a shooting guard. How many assignments are possible?
Solution: There are 5! arrangements, and there are 3 labels – {P}, {S}, and {N,N,N}. Thus there are 5! / (1! 1! 3!) = 20 outcomes.