Question 1193292: 13 people want to play a game. There can be any number of people on each team but only 2 teams total. How many combinations of teams can be formed?
I need steps and answer using nCr formula.
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
The statement of the problem does not make it absolutely clear; but I will assume each of the 13 people is on one team or the other.
So to form the teams, we can choose some number out of the 13 to be on one team, and the ones not chosen form the other team.
For example, if the first team chosen has 5 players, then the number of ways of forming the two teams is 13C5.
Then, since the first team chosen can have any number of players, the total number of ways of forming the two teams is
13C0 + 13C1 + 13C2 +...+ 13C12 + 13C13 = 2^13 = 8192
However, presumably a "team" with 0 players would not be considered a team, so the requirement of "2 teams total" would not be satisfied. So the number of ways of forming 2 teams from 13 players (eliminating the cases where the first team is made up of either 0 or 13 players) is
13C1 + 13C2 +...+ 13C12 = 8192-2 = 8190.
ANSWER: 8190
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Label the two teams as A and B.
Imagine we have 13 slots to represent the players. Each slot will be labeled either A or B.
Since we have 13 slots, and two choices per slot, this gives 2^13 = 8192 different ways to assign each player to a certain team. You can think of each slot like a binary light switch that turns on or off (for team A or team B).
If we require that each team needs 1 or more players, then we'd have to eliminate these cases:- Everyone going to team A (since team B wouldn't have anyone)
- Everyone going to team B (since team A wouldn't have anyone)
That means we've eliminated 2 cases to drop the count to 8192-2 = 8190
As you can see, there isn't a need to involve the nCr combination formula.
Though as the other tutor @greenestamps indirectly pointed out, summing all the terms along any row of Pascal's Triangle will yield a power of 2. Terms inside the triangle are directly connected to the nCr combination formula.
Answer: 8190
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