vela said:I think that number's way too small. The number of ways you can form just one team is already [C(10,3)*C(10,2)]*2=5400*2=10800. (The factor of 2 is there for this example because you can have 3 boys and 2 girls or 2 boys and 3 girls.)
Try working out what you get if you had 3 boys and 3 girls and had to form three teams with at least one boy and one girl each. It's easy enough to write down all the combinations and see how to count them.
vela said:No, that factor of two is only there because forming only one team leaves students that weren't chosen. When you form four teams, every student gets assigned to a team. If you count the ways to form all four teams, you end up covering every possible combination, so there's no need to double the number.
For instance, suppose you had three girls (A, B, and C) and three boys (0, 1, and 2), and you want to form two teams, each with at least one boy and one girl. One possible team is AB0, so the other team has to be C12. When you count AB0, you're already counting C12. If you were to count the team C12 again separately, you'd be double-counting.
vela said:Your calculating the ways you can form the first team, the second team, and the third team correctly. You're just not calculating the total number of possibilities correctly.
Think about flipping 3 coins. Your calculation is akin to saying since the each coin can come up heads or tails, there are 2+2+2=6 ways possible outcomes of flipping 3 coins, but there's actually 8: hhh, hht, hth, htt, thh, tht, tth, and ttt.
Mandark said:In your edit you should be dividing by 4 rather than multiplying because the two boy teams (teams with 3 boys) can be switched and it wouldn't make a difference to the partition, same goes with the two girl teams (3 girls on the team).
Mandark said:In your edit you should be dividing by 4 rather than multiplying because the two boy teams (teams with 3 boys) can be switched and it wouldn't make a difference to the partition, same goes with the two girl teams (3 girls on the team).
Remember, you're counting possible partitions, not possible teams. As Mandark noted, you can think of forming two boy teams and two girl teams. Every partition will contain a pair of 3B-2G teams and pair of 3G-2B teams. If you count the 3B-2G teams separately from the 3G-2B teams, you will be counting each partition that contains a specific combination of those teams multiple times.Roni1985 said:Now that I am reading it again, why did we need just to divide by 4 and not also multiply by ?
I think Vela's method was correcct
There are 2 options how we can form the first group 2*[C(10,3)*C(10,2)]- it can either be 3B 2G OR 3G 2B. therefore, we need to multiply by two ...
I guess we need to multiply the numerator by 8 ?! and divide by 4 ? (I also understand why you are saying that we need to divide and I think you are right about this. But I am not sure why not to multiply ?!)
Mandark said:That reasoning is correct for the first group, but try to continue it for the 2nd and 3rd groups (particularly the 3rd), and you'll begin to see where the difficulty lies in that approach (you won't end up with the same product as you wrote).
Giving an interpretation to the product you wrote down, you can think of the process as selecting a boy group, then another boy group, then a girl group, then another girl group (here boy group means group containing 3 boys). How do you select the first boy group? C(10,3) * C(10,2). Once this group is selected how do you select the second boy group? C(7,3) * C(8,2). Then the first girl group C(4,2) * C(6,3) and finally the last girl group C(2,2) * C(3,3). Once you do this you've over counted by a factor of 4, since you've counted the boy groups A then B, and B then A as different. Same goes for the girl groups.
vela said:Remember, you're counting possible partitions, not possible teams. As Mandark noted, you can think of forming two boy teams and two girl teams. Every partition will contain a pair of 3B-2G teams and pair of 3G-2B teams. If you count the 3B-2G teams separately from the 3G-2B teams, you will be counting each partition that contains a specific combination of those teams multiple times.
Again, to make what we're saying more concrete, I encourage you to work out what happens if you have fewer boys and girls and form smaller teams. You can list every possible team and possible partitioning and count them. Then try to calculate those numbers mathematically, and you'll hopefully see how the calculations work out.
The "Basketball roster probability problem" is a mathematical problem that involves determining the number of possible combinations for creating a basketball roster from a pool of players with different positions and skills.
The "Basketball roster probability problem" is important because it helps coaches and team managers make strategic decisions when selecting players for a team. It also allows for the optimization of team composition based on the strengths and weaknesses of individual players.
The "Basketball roster probability problem" can be solved using the principles of combinatorics and probability. It involves calculating the total number of possible combinations for a given number of players and positions, and then determining the probability of each combination occurring.
The main factors that affect the "Basketball roster probability problem" are the number of players available, the number of positions to be filled, and the specific skills and abilities required for each position. Other factors may include team chemistry, player injuries, and coaching strategies.
The principles of the "Basketball roster probability problem" can be applied to other team sports such as soccer, football, and baseball. It can also be used in other areas such as employee scheduling, project management, and event planning.