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A group of six friends want to play paintball game. In how many

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Bunuel
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A group of six friends want to play paintball game. In how many [#permalink] New post   04 Dec 2022, 08:42
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A group of six friends want to play paintball game. In how many different ways can the group be divided into 3 teams of 2 people each?

(A) 6
(B) 15
(C) 20
(D) 45
(E) 90

 


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Bunuel
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Re: A group of six friends want to play paintball game. In how many [#permalink] New post   04 Dec 2022, 08:45
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Bunuel wrote:
A group of six friends want to play paintball game. In how many different ways can the group be divided into 3 teams of 2 people each?

(A) 6
(B) 15
(C) 20
(D) 45
(E) 90

 


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Similar question to practice: https://gmatclub.com/forum/in-how-many- ... 02951.html Do you see the difference between these two ?
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Re: A group of six friends want to play paintball game. In how many [#permalink] New post   04 Dec 2022, 09:26
Bunuel wrote:
A group of six friends want to play paintball game. In how many different ways can the group be divided into 3 teams of 2 people each?

(A) 6
(B) 15
(C) 20
(D) 45
(E) 90

 


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The groups are identical. The number of ways of arranging "n" people into "y" identical groups consisting of "x" members in each group is given by -

\(\frac{n ! }{ (x!)^y * y!} \)

n = y * x

In this case n = 6
y = 3
x = 2

Number of ways = \(\frac{6! }{ (2!)^3 * 3!}\) = 15

Option B
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Re: A group of six friends want to play paintball game. In how many [#permalink] New post   04 Dec 2022, 09:41
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gmatophobia wrote:
Bunuel wrote:
A group of six friends want to play paintball game. In how many different ways can the group be divided into 3 teams of 2 people each?

(A) 6
(B) 15
(C) 20
(D) 45
(E) 90

 


Enjoy this brand new question we just created for the GMAT Club Tests.

To get 1,600 more questions and to learn more visit: user reviews | learn more

 



The groups are identical. The number of ways of arranging "n" people into "y" identical groups consisting of "x" members in each group is given by -

\(\frac{n ! }{ (x!)^y * y!} \)

n = y * x

In this case n = 6
y = 3
x = 2

Number of ways = \(\frac{6! }{ (2!)^3 * 3!}\) = 15

Option B


:thumbsup: How would you solve if you did not know the formula?
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gmatophobia
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Re: A group of six friends want to play paintball game. In how many [#permalink] New post   04 Dec 2022, 09:55
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Bunuel wrote:
:thumbsup: How would you solve if you did not know the formula?


Three teams need to be formed as shown.

_ _ _

We have six students, two students can be selected in 6C2 ways.

As two students are selected, from the remaining four we can select another set of two students in 4C2 ways.

The last set can be selected in 4C2 ways.

So total ways of selecting = 6C2 * 4C2 * 2C2 ways = \(\frac{6*5 }{ 2*1} * \frac{4*3}{2*1} * \frac{2 * 1}{2*1\\
}\) = \(\frac{6! }{ (2!)^3}\)

However note, that the three groups/teams are identical hence the arrangement doesn't matter so we have to de arrange them in \(\frac{1 }{ 3!}\) ways

Total = \(\frac{6! }{ (2!)^3}\) * \(\frac{1}{3!}\) = 15 ways
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Re: A group of six friends want to play paintball game. In how many [#permalink]
04 Dec 2022, 09:55
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