A teacher wants to select a team of 5 players from a group

Question

A teacher wants to select a team of 5 players from a group of 9 players. However, she needs to keep the following constraints in mind: If Jane is in the team, Sue should also be included in the team and vice versa.

In how many ways can the teacher select the team for a tournament?:

A) 21
B) 35
C) 56
D) 120
E) 126

In how many ways can the teacher select the team for a tournament?:

A) 21
B) 35
C) 56
D) 120
E) 126

Gladiator59 · Accepted Answer

Based on the constraints there could be two types of teams:
1. One with both Jane and Sue.
2. One without either.

1. # Of ways to select team with both Jane and Sue.

= 7C3 ( Select the other three team members as Jane and Sue are a given)
= 35.

2. # of ways to select a team without Jane and Sue.

= 7C5
= 21.

Total numbers of teams possible is 1. + 2.
Hence # teams = 35 + 21 = 56.

Hence Option (C) is our answer.

Best,
Gladi

Wildflower · Answer

Without any restriction we can select 5 players from a team of 9 by --> 9C5 = 9!/(4!5!) = 126 ways.

Restriction:  If Jane is in the team, Sue should also be in the team.  So we'll treat Jane and Sue as one group, and between them they can arrange in 2! ways = 2 ways.

Now, we need to select 3 more players out of 7 (as Jane and Sue are already in the team), which can be done by 7C3 = 7!/(4!*3!) = 35 ways.

Total = 35*2 = 70 ways

Total --> 126 - 70 = 56 ways.

houston1980 · Answer

There are 2 ways to arrange the team

First option.

Jane and Sue are selected in the team. Therefore we have to choose 3 additional team members from 7 = (9-(Jane+Sue)) = 9-2

7!/(4!3!) = 7*6*5/3! = 35

Second option.

Jane and Sue are outside the team. Therefore we have to choose 5 team members from 7 = (9-(Jane+Sue)) = 9-2

7!/(5!2!) = 7*6/2! = 21

Our answer is addition of two options above 35+21 = 56.

Hence    option C = 56    is the answer.

PPhoenix77 · Answer

there will be two cases: 1. Jane and sue are in the team
2. Jane and sue are not in the team

Case 1. 3 players need to be selected from remaining 7 = 7C3 =35
Case 2. 5 players need to be selected from remaining 7= 7C5= 21

so, total no. of ways = 35+21= 56
Correct Answer= C

EgmatQuantExpert · Answer

Solution

Given: 
• A group contain 9 players. 
• Jane and Sue are 2 such players in the group.

To find:  
• The number of ways in which the teacher can select a team of 5 players from the 9 players in the group.

Approach and Working out:  
• The only constraint given in the question is that the team either consists both Jane and Sue, or, consist none of them. 
 o Thus, it is better to solve the sum keeping the two conditions in mind.

Case 1:  
 When both Jane and Sue are in the team.  
In this case, two out of the five players are already in the team. 
Thus, we need to select 3 more players. 
The total number of players from which we can select 3 = 9 -2(Jane and Sue) = 7
Ways to select 3 players from 7 players =  ^7c_3 .

Case 2:  
 When neither Jane nor Sue is in the team.  
In this case, we need to select 5 players from 7 players. (Because we cannot consider Jane and Sue in the team)
Ways to select 5 players from 7 players =  ^7c_5

Since both the above cases can be true, we will add them to get the final answer. 
Ways to choose the team =  ^7c_3  +  ^7c_5  = 56

Hence, the   correct answer is option C.

EgmatQuantExpert · Answer

Alternate Solution

Given:  
• A group contain 9 players. 
• Jane and Sue are 2 such players in the group.

To find:  
• The number of ways in which the teacher can select a team of 5 players from the 9 players in the group.

Approach and Working out:  
• If either one of Jane or Sue is in the team, our given condition will be void. 
 o If we subtract these cases from the total possible cases, we will reach the answer.  
• Thus, we can write:
 o Total Possible selections = Total selections (Without Constraint) – Total selections where either Jane or Sue, but not both is in the team.  
 Note: When we are selecting either Jane or Sue, we must select 4 players from the remaining 7 players (9 – Jane and Sue) and 1 from Jane and Sue.  
Total Selections (Without Constraints) =  ^9c_5  = 126. 
Total selections where either Jane or Sue, but not both is in the team =  ^7c_4  *  ^2c_1  = 70
Total required selections = 126 – 70 = 56.

Hence, the   correct answer is option C.

ScottTargetTestPrep · Answer

If both Sue and Jane do make the team, the number of ways to select the team is 7C3 since we need to select 3 more players from the remaining 7 players:

7C3 = 7!/(3! x 4!) = (7 x 6 x 5)/3! = 35 ways

If both Sue and Jane do not make the team, the number of ways to select the team is 7C5:

7C5 = (7 x 6 x 5 x 4 x 3)/(5 x 4 x 3 x 2) = 7 x 3 = 21 ways

So the total possible ways is 35 + 21 = 56.

Answer: C

EMPOWERgmatRichC · Answer

Hi All,

We're told that a teacher wants to select a team of 5 players from a group of 9 players. However, she needs to keep the following constraints in mind: If Jane is in the team, Sue should also be included in the team and vice versa. We're asked for the number of different ways that the teacher can select the team for a tournament. This question is a Combination Formula question with a 'twist': you either need BOTH girls on the team OR you need NEITHER girl on the team. Those two options require slightly different calculations.

Combination Formula = N!/K!(N-K)! where N is the total number of people and K is the size of the subgroup.

For Jane and Sue to be on the team, those two players would take 2 of the 5 'spots', so the other 3 spots would be chosen from the remaining 7 players... 
7!/3!(7-3)! = (7)(6)(5)/(3)(2)(1) = 35 options

For NEITHER Jane NOR Sue to be on the team, the 5 'spots' would be chosen from the other 7 players... 
7!/5!(7-5)! = (7)(6)/(2)(1) = 21 options

Total possible teams = 35 + 21 = 56

Final Answer:  C

GMAT assassins aren't born, they're made, 
Rich

harshchougule · Answer

Can anyone explain what am i doing here?
Since Jane and sue are to be includes I treat them as one

so we have to find number of ways to form group of 5 from 8 members
i.e 8C5 = 56

I know this is wrong but dont know how.

Bunuel · Answer

If you treat them as one unit, then if this unit is chosen with 4 other units, you would end up with 6 people in the group: Jane and Sue, plus 4 others. When they are not included, you would get any 5 people except Jane and Sue. Therefore, you should split the cases as shown in the solutions above.

RAHUL208 · Answer

IF Jane is selected sue should also be selected(Jane then sue ), but how does it mean if Jane is not selected sue cannot be part of the team.
1) Jane is selected (Jane, Sue ) 7C3 = 35
2) Jane is not selected 8C5 = 56

35 + 56 = 91

Is this incorrect if yes how?

Krunaal · Answer

"If Jane is in the team, Sue should also be included in the team and  vice versa ."  - thus, if Sue is selected then Jane needs to be there => which also means, you cannot keep Jane out and select Sue.

luisdicampo · Answer

Deconstructing the Question

We need to choose a team of 5 from 9 players.

The condition says that  Jane and Sue must either both be selected or both be excluded .
So the only valid cases are:
Jane and Sue both in, or Jane and Sue both out.

Step-by-step

If  Jane and Sue are both on the team , then 2 spots are already filled, so we must choose 3 more players from the remaining 7.

\frac{7!}{3!4!} = \frac{7\cdot 6\cdot 5}{3\cdot 2\cdot 1} = 35

If  Jane and Sue are both not on the team , then all 5 players must be chosen from the remaining 7.

\frac{7!}{5!2!} = \frac{7\cdot 6}{2\cdot 1} = 21

Total valid teams:

35 + 21 = 56

Answer: 56

A teacher wants to select a team of 5 players from a group

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