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Nine highschool boys gather at the gym for a game of minivolleyball
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24 Oct 2015, 01:27
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Nine highschool boys gather at the gym for a game of minivolleyball. Three teams of 3 people each will be created. How many ways are there to create these 3 teams? A) 27 B) 51 C) 90 D) 175 E) 280
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Nine highschool boys gather at the gym for a game of minivolleyball
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24 Oct 2015, 02:07
ShristiK wrote: Nine highschool boys gather at the gym for a game of minivolleyball. Three teams of 3 people each will be created. How many ways are there to create these 3 teams?
A) 27 B) 51 C) 90 D) 175 E) 280 Ans: ELet us give the name of 3 team to be formed. Team A, B, and C. We have been given that there are 9 people. 1) Number of ways selecting 3 people from 9 for Team A = 9C3 = 84 2) Number of ways selecting 3 people from remaining 6 for Team B = 6C3 = 20 3) Number of ways selecting 3 people from remaining 3 for Team C = 3C3 = 1 Now we got Ans1 = 9c3 * 6C3 * 3C3 = 1680 (Total arrangement of 3 teams, including repeated value of teams i.e ABC, ACB..and so on.....) Now, we got the total ways of selecting 3 team of 3 people from 9 people. But we have repeated the sequence for the team. That is this has included the ABC, ACB, CAB and so on..i.e 3!(Number of arrangement for 3 things.) Hence Ans = Ans1 / 3! = 280+1 kudos if it helped you
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Re: Nine highschool boys gather at the gym for a game of minivolleyball
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25 Oct 2015, 15:50
Number of ways in which n * g different items can be divided equally into g groups, each containing n items and the order of the group is not important, i.e. {ABC} is the same as {BCA}.
\(\frac{(n * g)!}{n!^g * g!}=\frac{(3 * 3)!}{3!^3*3!}=\frac{4*5*6*7*8*9}{2*3*2*3*2*3}=280\)



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Re: Nine highschool boys gather at the gym for a game of minivolleyball
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25 Oct 2015, 21:51
ShristiK wrote: Nine highschool boys gather at the gym for a game of minivolleyball. Three teams of 3 people each will be created. How many ways are there to create these 3 teams?
A) 27 B) 51 C) 90 D) 175 E) 280 Arrange the 9 boys in a straight line in 9! ways. The first three form team 1, next three form team 2 and last three form team 3. But in each team, the boys are arranged in first, second third positions so you need to unarrange them by dividing by 3! three times (once for each team). You get 9!/(3! * 3! * 3!) Also, there are no distinct teams  team1, team2 and team3. You just have three teams. So you also need to unarrange the three teams by dividing by another 3!. You get 9!/(3! * 3! * 3!) * 3! = 280 Answer (E)
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Re: Nine highschool boys gather at the gym for a game of minivolleyball
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26 Oct 2015, 01:56
Let us say the 9 members are 1,2,3,4,5,6,7,8,9 The first set of three could be selected in 9C3 ways. Now six members remain. The second set of three could be selected in 6C3 ways. The remaining three members would form the third team. Now after selection as above we would have the following 2 possibilities among other possibilities (1,2,3) (4,5,6) (7,8,9) (4,5,6) (1,2,3) (7,8,9) and so on But the above possibilities are not distinct since there is no ordering needed among the 3 teams selected. So we have to divide 9C3*6C3 by 3!=280 ways.
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Re: Nine highschool boys gather at the gym for a game of minivolleyball
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24 Dec 2015, 13:17
VeritasPrepKarishma wrote: ShristiK wrote: Nine highschool boys gather at the gym for a game of minivolleyball. Three teams of 3 people each will be created. How many ways are there to create these 3 teams?
A) 27 B) 51 C) 90 D) 175 E) 280 Arrange the 9 boys in a straight line in 9! ways. The first three form team 1, next three form team 2 and last three form team 3. But in each team, the boys are arranged in first, second third positions so you need to unarrange them by dividing by 3! three times (once for each team). You get 9!/(3! * 3! * 3!) Also, there are no distinct teams  team1, team2 and team3. You just have three teams. So you also need to unarrange the three teams by dividing by another 3!. You get 9!/(3! * 3! * 3!) * 3! = 280 Answer (E) Is this just another way of calculating (9C3*6C3*3C3)/3! or is it technically an entirely different approach/way of thinking about the problem?
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Re: Nine highschool boys gather at the gym for a game of minivolleyball
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27 Dec 2015, 23:35
redfield wrote: VeritasPrepKarishma wrote: ShristiK wrote: Nine highschool boys gather at the gym for a game of minivolleyball. Three teams of 3 people each will be created. How many ways are there to create these 3 teams?
A) 27 B) 51 C) 90 D) 175 E) 280 Arrange the 9 boys in a straight line in 9! ways. The first three form team 1, next three form team 2 and last three form team 3. But in each team, the boys are arranged in first, second third positions so you need to unarrange them by dividing by 3! three times (once for each team). You get 9!/(3! * 3! * 3!) Also, there are no distinct teams  team1, team2 and team3. You just have three teams. So you also need to unarrange the three teams by dividing by another 3!. You get 9!/(3! * 3! * 3!) * 3! = 280 Answer (E) Is this just another way of calculating (9C3*6C3*3C3)/3! or is it technically an entirely different approach/way of thinking about the problem? They are two different ways of thinking: 1. Out of 9 boys, select 3 in 9C3 ways to make group 1. Out of remaining 6, select 3 in 6C3 ways to make group 2. Then you have 3 remaining and you select 3 out of 3 in 3C3 ways to make group 3. But mind you, you don't have a group 1, group 2 and group 3 so to unarrange, you divide by 3! You get (9C3*6C3*3C3)/3! 2. Arrange all 9 boys in a row in 9! ways. First 3 boys are group 1, next 3 are group 2 and last 3 are group 3. The first 3 boys are arranged so unarrange them by dividing by 3!. The next 3 boys are arranged so unarrange them by dividing by 3!. The last 3 boys are arranged so unarrange them by dividing by 3!. Again, you don't have a group 1, group 2 and group 3 so to unarrange, you divide by 3! You get 9!/(3! * 3! * 3!) * 3!
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Re: Nine highschool boys gather at the gym for a game of minivolleyball
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24 Aug 2018, 04:09
ShristiK wrote: Nine highschool boys gather at the gym for a game of minivolleyball. Three teams of 3 people each will be created. How many ways are there to create these 3 teams?
A) 27 B) 51 C) 90 D) 175 E) 280 The first person selected must be combined with a pair formed from the remaining 8 people to create a team of 3. From the 8 remaining people, the number of ways choose 2 = 8C2 = (8*7)/(2*1) = 28. Since 3 of the 9 people have been used to form the first team, 6 people are left. The next person selected must be combined with a pair formed from the remaining 5 people to create a team of 3. From the 5 remaining people, the number of ways choose 2 = 5C2 = (5*4)/(2*1) = 10. Since 6 of the 9 people have been used to form the first 2 teams, 3 people are left. The next person selected must be combined with a pair formed from the remaining 2 people to create a team of 3. From the 2 remaining people, the number of ways choose 2 = 2C2 = (2*1)/(2*1) = 1. To combine our options for the 3 teams, we multiply: 28*10*1 = 280.
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Re: Nine highschool boys gather at the gym for a game of minivolleyball &nbs
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