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Bunuel
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

:thumbsup: How would you solve if you did not know the formula?
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Bunuel

: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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Oh I see. In the other question there was distinction between groups - Physics, Maths and History. Here there is no such distinction. So here needs divide by 3!.

Great detail and grouping of questions to show distinction.
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let the 6members be: A B C D E F

We have six students, two students can be selected in 6C2 ways for the first group
And 4C2 ways for the second group and 2C2 for the third.

So total ways of selecting = 6C2 * 4C2 * 2C2 ways = 15*6*1 = 90

but, here the groups are identical, AB in group one is similar to AB in group2 or in group3, so we must divide by 3!

therefore, 90/3! = 15

Option B
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