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

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by VJesus12 » Mon Apr 08, 2019 5:23 am

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C

D

E

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Difficulty—

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

[spoiler]OA=D[/spoiler]

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by [email protected] » Thu Apr 11, 2019 9:59 am
Hi VJesus12,

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

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by deloitte247 » Thu Apr 25, 2019 11:15 pm
Picking 5 players excluding (Jane and Sue ) out of the remaining 7 players from the team
$$=7_{C_5}$$
Picking 3 Players (Considering Jane and Sue) are already in the team out of 7 remaining players to form the team
$$7_{C_3}$$
$$7_{C_5}=\frac{7!}{5!\left(7-5\right)!}=\frac{7!}{5!2!}=\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{\left(5\cdot4\cdot3\cdot2\cdot1\right)\left(2\cdot1\right)}$$ $$=\frac{7\cdot6}{2}=\frac{42}{2}=21$$
$$7_{C_3}=\frac{7!}{3!\left(7-3\right)!}=\frac{7!}{3!4!}=\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{\left(3\cdot2\cdot1\right)\left(4\cdot3\cdot2\cdot1\right)}$$ $$=\frac{7\cdot5}{1}=35$$
$$7_{C_5}+7_{C_3}=21+35=56$$

$$Answer\ is\ Option\ C$$
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by swerve » Fri Apr 26, 2019 7:13 am
Case 1: Either of Jane and Sue is selected.

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

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

Therefore, total ways: case1+ case 2= 35+21=56 \(\Rightarrow\) __C__
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