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In how many different ways can a group of six students be equally

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Bunuel
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In how many different ways can a group of six students be equally [#permalink] New post   04 Dec 2022, 08:42
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In how many different ways can a group of six students be equally distributed into three classes: physics, mathematics and history ?

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

 


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Bunuel
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Re: In how many different ways can a group of six students be equally [#permalink] New post   04 Dec 2022, 08:45
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Bunuel wrote:
In how many different ways can a group of six students be equally distributed into three classes: physics, mathematics and history ?

(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

 



Similar question to practice: https://gmatclub.com/forum/a-group-of-s ... 02952.html Do you see the difference between these two ?
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gmatophobia
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Re: In how many different ways can a group of six students be equally [#permalink] New post   04 Dec 2022, 09:27
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Bunuel wrote:
In how many different ways can a group of six students be equally distributed into three classes: physics, mathematics and history ?

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


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

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

n = y * x

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

Number of ways = \(\frac{6! }{ (2!)^3 }\) = 90

Option E
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Bunuel
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Re: In how many different ways can a group of six students be equally [#permalink] New post   04 Dec 2022, 09:41
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gmatophobia wrote:
Bunuel wrote:
In how many different ways can a group of six students be equally distributed into three classes: physics, mathematics and history ?

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


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

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

n = y * x

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

Number of ways = \(\frac{6! }{ (2!)^3 }\) = 90

Option E


:thumbsup: How would you solve if you did not know the formula?
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gmatophobia
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Re: In how many different ways can a group of six students be equally [#permalink] New post   04 Dec 2022, 09:49
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Bunuel wrote:
:thumbsup: How would you solve if you did not know the formula?


There are 6 students, we can select two students for physics class in 6C2 ways.

As two students have been selected we have 4 students left, two student can be selected for mathematics class in 4C2 ways

As four students have been selected we have two students who can be selected for history class in 2C2 ways

Total = 6C2 * 4C2 * 2C2 = \(\frac{6*5 }{ 2*1} * \frac{4*3}{2*1} * \frac{2 * 1}{2*1\\
}\) = \(\frac{6! }{ (2!)^3}\)

Total = 90 ways
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Re: In how many different ways can a group of six students be equally [#permalink]
04 Dec 2022, 09:49
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