3 teams each with 5 basket players. How many combinations...

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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There are 3 teams each with 5 basket players. How many combinations are there if we were to pick 2 players from the 3 teams such that no team was picked twice?


A) 50
B) 23
C) 75
D) 28
E) 45

I know the answer is (3C2)*(5C1)*(5C1)=75
What other methods could we employ to find the same answer??

One method I am trying but its not working is: So we have 3 teams 5 each, so 15 players. I label each player by number and by team. So teams A,B,C. and the players are 1_a, 2_a,3_a,4_a,5_a, 1_b, 2_b ... 5_b, 1_c, 2_c .. 5_c.

We have a two slots. 15*14=190 permutations. But some of them are duplicates and order doesn't matter (eg. {3_a,5_c}={5_c,3_a}). So we divide by the number of slots 2!. 190/2!=95. Now we also have another constraint "No two players can be chosen from the same team". How many many ways are two 2 players from the same team are chosen? From team A we have 5 players. 5*4 ways to move them around. But again order doesnt matter so we divide by the number of slots 2!. So 10 ways. Same thing for team B and C. So we have a total of 30 ways to pick two player from the same team. from the 95 I subtract 30 to get 65. What am I doing wrong?

Are there any other ways to solve this problem other than what I laid out here?
Source: — Quantitative Reasoning |

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by [email protected] » Sun Sep 01, 2013 12:51 pm
Hi johndoe88,

You don't need to do any complex math to solve this question. Here's how:

Team 1: A, B, C, D, E

Team 2: 1, 2, 3, 4, 5

Team 3: V, W, X, Y, Z

Since you're only picking 2 players and they can't both be from the same team, the options are limited:

Take a player from Team 1: "A" and you can pair this player with 10 other players (anyone from Team 2 or Team 3)

So, 5 players on Team 1 = (5)(10) possible pairings = 50 possibilities

Now, everyone on Team 2 has ALREADY been paired with everyone on Team 1, so all that's left is to pair players from Team 2 with players from Team 3. That would be (5)(5) = 25 additional possibilities.

Since everyone from Team 3 has ALREADY been paired with everyone from Team 1 and from Team 2, there's nothing else to add.

Total Possibilities = 50 + 25 = 75

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Rich
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by johndoe88 » Sun Sep 01, 2013 2:54 pm
That is an interesting way of solving it Rich. Well can you elaborate about a little about the approach I took. What is wrong with y thinking?

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by [email protected] » Sun Sep 01, 2013 10:26 pm
Hi johndoe88,

Your logic is fine, although it's a bit long-winded. The reason why it didn't get you the correct answer is because you made a silly math mistake:

(15)(14) = 210

Your original calculation had it equaling 190.

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by johndoe88 » Mon Sep 02, 2013 1:32 pm
Rich, I really appreciate your help and patience. I am trying to understand another concept that is confusing me with this problem and other combinations problems.

My problem is trying to understand the difference between the two methods I am going to outline below.

Solution 1) I have 3 teams and I pick 2 of those teams for my 2 slots (3C2). From the 2 teams I pick 1 each from their 5 players(5C1)*(5C1). So answer is (3C2)(5C1)(5C1)=75

Solution 2) I have 3 teams and I pick 1 team for the first slot. From that chosen team I pick 1 player out of the 5 (5C1). For my 2nd slot, from the remaining 2 teams, I pick another (2C1). From that chosen team I pick 1 player out of the 5 (5C1).
Solution (3C1)(5C1)(2C1)(5C1)=150.
This solution isn't correct. I have to divide by 2! because of double counting but ????? why?????. What clue tells you are double counting?

I am trying to understand why I in the first problem I don't have to divide by 2! and what in the second problems mandates me that I divide by 2!.
How do you tell if your double counting?? Is it because in the second problem. I literally went slot by slot filling the values in order? What is causing the double counting?

I really appreciate your help Rich.

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by Brent@GMATPrepNow » Sun Sep 08, 2013 6:11 am
johndoe88 wrote:There are 3 teams each with 5 basket players. How many combinations are there if we were to pick 2 players from the 3 teams such that no team was picked twice?

A) 50
B) 23
C) 75
D) 28
E) 45
I know that johndoe88 already said that one approach results in the answer: (3C2)(5C1)(5C1) = 75. I thought I'd cover the rationale for this answer in greater detail.

The above solution applies something called the Fundamental Counting Principle (FCP). For more information about the FCP, see our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775

We're going to take the task of selecting the 3 players and break it into stages.

Stage 1: Choose the 2 teams from which we will select the players.
Since the order of the selected teams does not matter, we can use combinations.
We can select 2 teams from 3 teams in 3C2 ways (3 ways).
If anyone is interested, we have a free video on calculating combinations (like 3C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Stage 2: Choose a player from one of the selected teams.
There are 5 players to choose from, so we can complete this stage in 5 ways

Stage 3: Choose a player from the other selected team.
There are 5 players to choose from, so we can complete this stage in 5 ways

By the Fundamental Counting Principle, we can complete all 3 stages (and thus select 2 players) in (3)(5)(5) ways ([spoiler]= 75 ways[/spoiler])

Answer = C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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