# Combinatorics problem - Amount of basketball teams

The question goes as follow:

A basketball team of five players must be selected from among $$12$$ players. The players are first divided into two groups of six in each, of which you must choose three from group $$1$$ and two from group $$2.$$

a) How many different teams can be selected from among the $$12$$ players.

b) How many teams do not contain both the weakest and the strongest player?

My solution for a):

Choose $$3$$ from group $$1$$: $$\binom63$$(sorry I don't know how to do the symbol)

Choose 2 from group 2: $$\binom62$$

Total: $$\binom63 \times \binom62 = 300$$ teams.

My solution for b):

Split into cases:

1.) Both players in group 1: $$\binom{4}{3}\times\binom62=60$$ teams

2.) Both players in group 2: $$\binom63 \times \binom42 =120$$ teams

3.) Players in different groups: $$\binom53 \times\binom52=100$$ teams

Total: $$60 + 120 + 100 = 280$$ teams

Basically I got it wrong for both a) and b) and I can't seem to figure out why. Any solution or help to this problem?

• The way that I would interpret the problem statement is that you are allowed to divide the 12 players into two groups of 6 in any way that you like, not just a single specified way. In this case, the initial step of dividing the team in half is a red herring and the solution to (a) is 12 choose 5. That's just my interpretation though. Commented Dec 15, 2022 at 19:47
• You can produce $\binom{n}{k}$ by typing $\binom{n}{k}$. This MathJax tutorial explains how to typeset mathematics on this site. Commented Dec 16, 2022 at 1:07

There are , unfortunately, various ways to interpret the question. The interpretation I am taking (based on mention of Group 1 and Group2) is that the 12 are divided, once for all, into two distinct groups (eg people being allotted hostels, for instance) and that remains fixed.

In other words, if A,C,D,F,G,H have benn allotted to Group 1, that's it, you can't say that Group 1 could instead have been A,B,C,D,E,F

On this premise, you can't ignore the initial division into groups.

Ways to form Gr $$1 = \binom{12}{6}$$, Gr $$2$$ automatically formed

(a): Ways to form teams $$= \binom{12}6\binom63\binom62 =277 200$$

(b): A better strategy would be to exclude from the total computed above, teams having both the "marked" players, dividing into cases as you have done

• I think this is still not correct. If ${12 \choose 5} = 792$, then how can there be 277200 ways to form teams? Given your interpretation of the problem, I would say that leun did the problem correctly and that the solution is ${6\choose 3}{6 \choose 2} = 300$. You would not need to multiply by ${12 \choose 6}$ if the initial groups are set in stone. Commented Dec 15, 2022 at 23:31
• @ChrisDuerschner: U agree with you that the. question is very poorly worded, But bear in mind that OP says that his answers (going by your interpretation) have been marked wrong. Commented Dec 16, 2022 at 3:43
• Thanks for the response. But basically on b) coudn't I do the same as you've done here? I just mulitply every case by $\binom{12}{6}$. And then add all the cases together as I've done. Also @ChrisDuerschner I agree that the task is poorly worded but basically we don't have an inital determined set of two groups and maybe thats why I got it wrong
– leun
Commented Dec 16, 2022 at 8:41
• Also by the way the answer seems extremely odd to me even though somehow the idea makes sense. $\binom{12}{5}$ should be the maxiumum amount of possible teams to ever create. I can't seem to understand how there could be 277200 ways to create this team
– leun
Commented Dec 16, 2022 at 9:20
• @leun: I know the whole thing appears strange, but given that both your answers were marked wrong, the only other possible idea that struck me was that a person is identified by name as well as the team, which means if Jane is on team $1$, she is $J1$, and if on team $2, J2$. (Note that this sidesteps the question of overcounting, too) Commented Dec 16, 2022 at 10:12

I believe that true blue anil's solution 'double counts' many teams. To see why, consider a simpler case of the problem with only a three player team being selected from a set of six players. Let's call the players A, B, C, D, E, and F. One way to choose a team would be to initially divide the players according to (A,B,C) and (D,E,F) then, choosing A and B from the first group and D from the second group, arrive at the team (A,B,D).

However you could also initially group the teams according to (A,C,D) and (B,E,F). Then, choosing A and D from the first group and B from the second, arrive at the same team as before.

If you are allowed to initially divide the 12 players at will, instead of being presented with an already chosen initial grouping, then the solution to (a) is just $${12 \choose 5}$$ - the total number of teams.

The other way I know that 277200 is not the correct answer to (a) is that it is greater than $${12 \choose 5} = 792$$. The number of possible teams that you get by initially grouping, then choosing 3 and 2 from the groups cannot be greater than the number of teams you would get by just choosing 5 players from the 12 without initial grouping.

EDITING again to address part (b) - note my previous edit was incorrect, and that this answer also labors under the assumption that the initial grouping is not predetermined.

So we need to subtract from the total number of teams the teams that have both of two specific players. The number of teams we should subtract in this case is $${10 \choose 3} = 120$$, so the solution would be 792-120=672. Why? Suppose we exclude the two players from the initial group of 12, leaving only 10 players left. Then, we need to count the number of teams we could create with only 3 players. Then, if we just tack the 2 excluded players onto these teams, we have every possible team with both of the excluded players. In general, if 'n' is the number of players to choose from, 'k' is the size of the teams, and 'a' is the number of players we need to exclude, then the number of teams that don't have the excluded players is $${n\choose k} - {n-a \choose k-a}$$.

• Maybe we should consider (A,B,C,D,E) is not similar to (B,C,D,E,A), because we choose Center , then Post-guard, then ... ... ; it is the only reason I can find if we say that answer 300 for Question 1 is wrong. For question 2, the constraint is not to exclude teams with Best or Worst, but to exclude Teams with both (Best+Worst). Commented Dec 16, 2022 at 12:23
• Yeah, I realized that I misunderstood part (b), so I deleted that part of my answer. As for caring about order... that seems a little far fetched, but I agree that if the initial order is predetermined then the answer should be 300. Commented Dec 16, 2022 at 14:45