Ten highschool boys gather at the gym for a game of basketball. Two te

Question

Ten highschool boys gather at the gym for a game of basketball. Two teams of 5 people each will be created. How many ways are there to create these 2 teams?

(A) 90
(B) 105
(C) 126
(D) 252
(E) 525

adiagr · Answer

Question involves simply selecting 5 players out of 10 players.

10C5 =  \frac{10!}{5!5!}  =  \frac{10.9.8.7.6}{5!}  = 252

Answer: D

Divyadisha · Answer

Number of ways to select 5 people out of 10= 10!/5!5!= 252

D is the answer

KS15 · Answer

I think C should be the answer here.

AbdurRakib · Answer

Please explain for you position.I'm confused to choose C

gagan0001 · Answer

For the first team 5 players can be selected as 10C5 = 252 ways
For second team 5 players can be selected as 5C5= 1 way
Now we have 2 teams and the total number of ways = (252*1)/2! = 126
I have divided by 2! because order in which the teams are created does not matter.

JD01 · Answer

This is a problem of combination.
- First we need to choose  5  boys from  10  =  C^5_{10} 
- Second we need to choose  5  boys from remaining  5  =  C^5_5  (This is because we have already picked  5  boys)
- Also as the order of team selection does not matter we can simply divide the whole equation by  2!  instead of  C^2_{10} 
 
 \frac{C^5_{10} * C^5_5}{2!}   =   \frac{\frac{10!}{5!*5!}*1}{2!}    =   126

Answer = C

MithilaGauri · Answer

I feel there is no need to divide by 2. We are only selecting 5 players out of 6??

Posted from my mobile device

MithilaGauri · Answer

I feel there is no need to divide by 2. We are only selecting 5 players out of 6??

Posted from my mobile device

MithilaGauri · Answer

I feel there is no need to divide by 2. We are only selecting 5 players out of 6??

Posted from my mobile device

pratik2018 · Answer

a,b,c,d,e,f,g,h,i,j be the 10 players

lets say a,b,c,d,e got selected once in 10C5 case so remaining team in 5C5 is f,g,h,i,j
at some point f,g,h,i,j will also be selected in 10C5 then remaining team in 5C5 will be a,b,c,d,e.

So there is a repetition here and thus we need to divide the total possible cases by 2!.

ScottTargetTestPrep · Answer

The first team (Team A) can be created in 10C5 ways:

10! / (5! x 5!) = (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2) = 3 x 2 x 7 x 6 = 6 x 42 = 252

The second team (Team B) can be selected in 5C5 = 1 way.

Since the order in which the two teams are listed (i. e., AB versus BA) does not matter, we must divide by 2!, so the total number of ways is 252/2 = 126.

Answer: C

bumpbot · Answer

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Ten highschool boys gather at the gym for a game of basketball. Two te

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