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There are 3 states and 3 students representing each state. [#permalink]

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14 Apr 2010, 05:29

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There are 3 states and 3 students representing each state. In how many ways can 5 students be chosen such that at Least one student is chosen from each state.

1) There are 3 states and 3 students representing each state. In how many ways can 5 students be chosen such that at least one student is chosen from each state.

To choose 5 students so that atleast one student will represent each state can be done in two ways:

A. 3-1-1 (3 students from 1 state and 1 student from other two states) 3C1*3C3*3C1*3C1=27 3C1 - # of ways to choose 3-student state; 3C3 - # of ways to choose 3 students from 3-student state; 3C1 - # of ways to choose 1 student from the first 1-student state; 3C1 - # of ways to choose 1 student from the second 1-student state.

OR B. 1-2-2 (1 student from 1 state and 2 students from other two states) 3C1*3C1*3C2*3C2=81 3C1 - # of ways to choose 1-student state; 3C1 - # of ways to choose 1 student from the 1-student state; 3C2 - # of ways to choose 2 students from the first 2-student state; 3C2 - # of ways to choose 2 students from the second 2-student state.

It's necessary that 1 student is selected from each state. So number of ways 1 student can be selected from 1 state = 3. So number of ways 1 student can be selected from each of the 3 state is 3*3*3 = 27.

3 of the 9 have now been selected and 6 remain; out of which 2 need to be selected. This can be done in 6C2 ways.

So the answer is 27*6C2 = 405 If the 5 people were to be selected for 5 different posts, then the answer would be 405*5!
_________________

In a Normal Distribution, only the Average'Stand Out'

It's necessary that 1 student is selected from each state. So number of ways 1 student can be selected from 1 state = 3. So number of ways 1 student can be selected from each of the 3 state is 3*3*3 = 27.

3 of the 9 have now been selected and 6 remain; out of which 2 need to be selected. This can be done in 6C2 ways.

So the answer is 27*6C2 = 405 If the 5 people were to be selected for 5 different posts, then the answer would be 405*5!

Above solution is not correct. There will be duplications in it. Let's see this on another example: two states 3 from each. We should choose 3 students so that atleast one will be from each state.

State 1: A, B, C State 2: X, Y, Z

According to the above solution answer should be 3C1*3C1*4C1=36.

But actual groups are: ABX; ABY; ABZ;

ACX; ACY; ACZ;

BCX; BCY; BCZ;

XYA; XYB; XYC;

XZA; XZB; XZC;

YZA; YZB; YZC.

Total 18 groups.

According to my solution: one pattern 1-2 --> 2C1(choosing which state will provide with one student)*3C1*(choosing this one student from 3)*3C2(choosing 2 student from another state)=2*3*3=18 (correct answer).

Total number of possibilities \(= 9C5 = 9C4 = \frac{9*8*7*6}{4*3*2} = 126\) From this we need to subtract the possibilities where the 5 students are chosen solely among the other 2 states. This can be done in \(3*6C5 = 3*6C1 = 18 ways\) So \(126 - 18 = 108 ways\)
_________________

In a Normal Distribution, only the Average'Stand Out'

Re: There are 3 states and 3 students representing each state. [#permalink]

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31 Aug 2014, 00:23

lprassanth wrote:

Total number of possibilities \(= 9C5 = 9C4 = \frac{9*8*7*6}{4*3*2} = 126\) From this we need to subtract the possibilities where the 5 students are chosen solely among the other 2 states. This can be done in \(3*6C5 = 3*6C1 = 18 ways\) So \(126 - 18 = 108 ways\)

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