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

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.

Gladiator59 has already provided the method I thought of, so I am late to this party!
Wildflower thank you for sharing a different perspective!

You're welcome! This is my usual approach

\(7C5\) -> Picking 5 players (excluding Jane & Sue) out of remaining 7 players to form the team.
\(7C3\)-> Picking 3 players (considering Jane & Sue are already there in the team) out of 7 remaining players to form the team.

\(7C5\)+\(7C3\)=21+35=56 (C).

Selecting 5 players from a group of 9 players:
Selected | Not selected
_ _ _ _ _ | _ _ _ _

1. Given that there is no constraints, there are 9C5 = \frac{9!}{(5!*(9-4)!)} = 126 ways to arrange.

2. Given that Jane and Sue has to be on the same team, calculate number of ways given if Sue is in the team, Jane is not in the team or vise versa.

_ _ _ _ J | _ _ _ S or _ _ _ _ S | _ _ _ J

Ignoring J and S as they are fixed, there are 2*(7C4) = \frac{7!}{(4!*(7-4)!)} = 70 ways.
_ _ _ _ | _ _ _

Subtracting 2 from 1, 126 Ways - 70 Ways = 56 Ways.

The answer is C) 56

Hello, dear friends! to be honest, I am not sure, but i think that the correct answer is E. Is it so or not?

Hey Therming

Can you please tell me why do you think that E is the answer?

Since this is a contest, you need to post the complete analysis of your solution to be eligible for the contest.

Awaiting your response.

Regards,
e-GMAT

Ans: C

Case 1: Either of Jane and Sue is selected.

In this case , both will be part of the team. Therefore, remaining 3 person can be selected from remaining 7 of the players in 7c3 ways. ie. 35 ways.

Case 2: Neither of 2 is elected. So, 5 player can be selected from remaining 7 people in 7c5 ways. i.e. 21 ways

Therefore, total ways: case1+ case 2= 35+21=56

The correct answer is option C.

Here is why:
This is a Permutation & Combination question, which can be solved easily if the basic concepts are clear.

In the question Jane and Sue need to be included in a team. So either 1) they are selected or 2) they aren't selected - no cases where only one of them is selected. Important to note, there is no importance of who is selected first. The number of ways will be the sum of both the ways both cases are selected -
1) Since the two are already selected, 3 teammates need to be selected from the remaining 7 i.e. 7C3 =35
2) In this case, the two can't be selected. So all 5 need to be selected from the remaining 7 i.e. 7C5 =21
Total = 35 +21 =56

Hence Option C

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

Wildflower - GREAT job on this one! PM me to get your GMAT Club tests!

Yay! Thank you!

Because Jane and Sue have to be chosen simultaneously.
So, there are 2 scenarios: they are chosen and they are not

If they are chosen, there are left with 3 choices in 7 people -> we have 7C3 = 35 ways to form a suitable team.
If they are not chosen, there are left with 5 choices in 7 people -> we have 7C5 = 21 ways

Combine 2 scenarios -> 35+ 21 = 56 -> C

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.

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.

5 team player need to be selected from 9 players

I) when Jane and Sue selected then 3 players need to be selected from 7
So 7C3=35

II)none from Jane and Sue selected

5 need to be selected from 7

So 7C5=21

Total= 35+21=56

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Please untag probability, Thank you.