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tough p n c [#permalink]

14 Apr 2010, 05:29

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

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Re: tough p n c [#permalink]

14 Apr 2010, 06:14

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jatt86 wrote:

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.

27+81=108.

Answer: 108.
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Re: tough p n c [#permalink]

15 Apr 2010, 22:23

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!
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Re: tough p n c [#permalink]

16 Apr 2010, 04:17

lprassanth wrote:

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).

Hope it's clear.
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Re: tough p n c [#permalink]

16 Apr 2010, 08:06

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DEAR MODERATOR,
not understood.

I think:
One from each of 3 states ( havin 3 st each) - 3c1*3c1*3c1 + 6c2 shd be the answer.
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Re: tough p n c [#permalink]

16 Apr 2010, 08:18

I think meghash3 is right. Thanks for pointing out.
I screwed it up by multiplying 27 and 6C2.
It should be 27 + 6C2

Could the moderator kindly point out the fault in this approach, if any?
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Re: tough p n c [#permalink]

16 Apr 2010, 13:56

5 slots to choose.

(3C2 * 3C2 * 3C1) --> 2 from state 1, 2 from state 2, 1 from state 3.
= 27.

We need to account for possibilities where we choose 1 from state 2, and 1 from state 1.
Therefore, 3*27 = 81.

3C3 * 3C1 * 3C1 --> 3 from state 1, and 1 each from the remaining states.
= 9.
As above, account for the other two possibilities.
Therefore, 3*9 = 27.

Total = 108.

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Re: tough p n c [#permalink]

17 Apr 2010, 05:55

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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Re: tough p n c [#permalink] 17 Apr 2010, 05:55

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