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Advanced Combinatorics Problem

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Advanced Combinatorics Problem [#permalink] New post 14 Feb 2007, 19:12
I have the following question.

Tanya bought 5 glasses for her kitchen - white, red, black, grey, and blue - and would like to display 3 of them on the shelf next to each other. If she decides that a black and a white glass cannot be displayed at the same time, in how many different ways can Tanya arrange the glasses?

I got the answer
Quote:
54
. Is it correct or not?
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Re: Advanced Combinatorics Problem [#permalink] New post 14 Feb 2007, 19:16
GMAT100 wrote:
I have the following question.

Tanya bought 5 glasses for her kitchen - white, red, black, grey, and blue - and would like to display 3 of them on the shelf next to each other. If she decides that a black and a white glass cannot be displayed at the same time, in how many different ways can Tanya arrange the glasses?

I got the answer
Quote:
54
. Is it correct or not?


5! - 4!*2! = 72
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Re: Advanced Combinatorics Problem [#permalink] New post 15 Feb 2007, 08:14
trivikram wrote:
GMAT100 wrote:
I have the following question.

Tanya bought 5 glasses for her kitchen - white, red, black, grey, and blue - and would like to display 3 of them on the shelf next to each other. If she decides that a black and a white glass cannot be displayed at the same time, in how many different ways can Tanya arrange the glasses?

I got the answer
Quote:
54
. Is it correct or not?


5! - 4!*2! = 72


IS the answer = 48

since White and Black cannot be there at the same time group them

now 4 glasses from which we need 3

4 C 3 * 2 (either black or white)

Now she can arrange 3 different glasses in 3! number of ways

3! * 4 C3 *2 = 6*4*2 = 48?
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 [#permalink] New post 15 Feb 2007, 09:58
48 is the answer for me as well. Assuming that either Black or White have to be on the shelf.

Thinking more...

Case 1 (White)
White, Blue, Red, Grey
4C3 ways = 4

Case 2 (Black)
Similar to Case 1, 4 ways

Case 3 (Neither Black nor White)

Now 3 three glasses can be selected in 3C3 ways = 1

Total ways 4+4+1 = 9

3 glasses can be arranged in 3! ways = 6

9*6 = 54

Now if the assumption was that either Black or White have to be on the shelf answer will be 48
4+4 = 8
8*6 = 48

Tricky question...
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 [#permalink] New post 15 Feb 2007, 10:23
budugu wrote:
amorpheus wrote:
48 is the answer for me as well. Assuming that either Black or White have to be on the shelf.

Thinking more...

Case 1 (White)
White, Blue, Red, Grey
4C3 ways = 4

Case 2 (Black)
Similar to Case 1, 4 ways

Case 3 (Neither Black nor White)

Now 3 three glasses can be selected in 3C3 ways = 1

Total ways 4+4+1 = 9

3 glasses can be arranged in 3! ways = 6

9*6 = 54

Now if the assumption was that either Black or White have to be on the shelf answer will be 48
4+4 = 8
8*6 = 48

when you did 4 c 3 it already includes case 3 ...! Hence 8 x 6

Tricky question...

what is the correct answer?
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 [#permalink] New post 15 Feb 2007, 10:29
budugu wrote:
amorpheus wrote:
48 is the answer for me as well. Assuming that either Black or White have to be on the shelf.

Thinking more...

Case 1 (White)
White, Blue, Red, Grey
4C3 ways = 4

Case 2 (Black)
Similar to Case 1, 4 ways

Case 3 (Neither Black nor White)

Now 3 three glasses can be selected in 3C3 ways = 1

Total ways 4+4+1 = 9

3 glasses can be arranged in 3! ways = 6

9*6 = 54

Now if the assumption was that either Black or White have to be on the shelf answer will be 48
4+4 = 8
8*6 = 48

when you did 4 c 3 it already includes case 3 ...! Hence 8 x 6

Tricky question...


Also what if we interpret that black or white have to be there then we have 8-3C3-3C3 = 6 ways, Hence 6+6 = 12*6 = 36
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 [#permalink] New post 15 Feb 2007, 13:09
I get 42, if this is wrong please explain why.

Total number of ways to sit 3 people out of 5 is = 5*4*3 = 60

There are 3 display locations, of which you want to display 2 (B and W) as well as one of the other 3. The total number of ways B and W can be displayed at the same time is (3c2) * 3=18.

I also did this drawing it out.

3*B*W
B*3*W
B*W*3

This gives 9, which needs to be doubled to 18 cause you can have BW or WB

If you subtract you get 60-18=42
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 [#permalink] New post 15 Feb 2007, 15:19
maxpowers wrote:
I get 42, if this is wrong please explain why.

Total number of ways to sit 3 people out of 5 is = 5*4*3 = 60

There are 3 display locations, of which you want to display 2 (B and W) as well as one of the other 3. The total number of ways B and W can be displayed at the same time is (3c2) * 3=18.

I also did this drawing it out.

3*B*W
B*3*W
B*W*3

This gives 9, which needs to be doubled to 18 cause you can have BW or WB

If you subtract you get 60-18=42


Here is my complete analysis:

We all agree that total ways to select 3 Glasses out of 5 is 60
B: Black
Bl: Blue
W: White
G: Grey
R: Red

Now lets take various cases:
Case1: B&W together
B W G
B W Bl
B W R
3 ways we can select and each can be arranged in 3! ways. Therefore total possible ways = 18----1

Case2: Either B or W
B G Bl
B G R
B R Bl
W G Bl
W G R
W R Bl
6 ways we can select and each can be arranged in 3! ways. Therefore total possible ways = 36----2

Case3: Neither B nor W
R G Bl
Only 1 way we can select and it can be arranged in 3! ways. Therefore total possible ways = 6----3

So if we add 1, 2, and 3 we get 60

So the answer can be 36 if either B or W have to be there but not both. I hope this is correct :)
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ANSWER is explained [#permalink] New post 15 Feb 2007, 17:07
The constraint combinatorics can be achieved in the following 2 ways:

FIRST WAY
=======
CHOOSE FIRST CONSTRAINT SCENARIO then OTHER

Here constraint is white and black
First choose black:
As you are selecting black, you cannot select white. It means that you have 4 glasses available to choose from
4 x 3 x 2 x 1= 24
Same with white:
Same as first
4 x 3 x 2 x 1= 24

Others
You are selecting neither white nor black
3 x 2 x 1 = 6

24 + 24 + 6=54

SECOND WAY
=========
Get ALL SCENARIOS and then subtract impossible scenario from the ALL SCENARIOS

All Scenarios
(5!)/(2!)=60

Impossible Scenarios
(3!)/(1!)=6

60-6=54
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 [#permalink] New post 16 Feb 2007, 07:27
The answer is 42

Total Outcomes 5*4*3 = 60

You have 3 spots, we need to eliminate where BW are both optiosn

so each time we have black and white in a position, there are 3 occurances of this since there are 3 other colors.

lets list outcomes with both black and white , note that color refers to the other 3 colors, so each one counts as 3


BW Color - 3
WB Color - 3

B Color W -3
W Color B -3

Color BW - 3
Color WB - 3

so adding up we get 18 subtracting fro m 60 60-18 = 42 ways
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 [#permalink] New post 16 Feb 2007, 07:36
terp26 wrote:
The answer is 42

Total Outcomes 5*4*3 = 60

You have 3 spots, we need to eliminate where BW are both optiosn

so each time we have black and white in a position, there are 3 occurances of this since there are 3 other colors.

lets list outcomes with both black and white , note that color refers to the other 3 colors, so each one counts as 3


BW Color - 3
WB Color - 3

B Color W -3
W Color B -3

Color BW - 3
Color WB - 3

so adding up we get 18 subtracting fro m 60 60-18 = 42 ways


In 42 ways you have the choices in which Neither B nor W are there... so I am not sure if that needs to be counted based on the question what you guys think?
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 [#permalink] New post 16 Feb 2007, 07:55
amorpheus wrote:
terp26 wrote:
The answer is 42

Total Outcomes 5*4*3 = 60

You have 3 spots, we need to eliminate where BW are both optiosn

so each time we have black and white in a position, there are 3 occurances of this since there are 3 other colors.

lets list outcomes with both black and white , note that color refers to the other 3 colors, so each one counts as 3


BW Color - 3
WB Color - 3

B Color W -3
W Color B -3

Color BW - 3
Color WB - 3

so adding up we get 18 subtracting fro m 60 60-18 = 42 ways


In 42 ways you have the choices in which Neither B nor W are there... so I am not sure if that needs to be counted based on the question what you guys think?


My reading of the question is that you can't have both B and W at the same time, but you can have 1 or the other, or none. So if you subtract out all the cases with B and W together you will get the answer.
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 [#permalink] New post 18 Feb 2007, 20:36
I would think that if they simply left out the restriction on the white and black pots, why wouldn't the answer be:5c3

you have 5 options, and you can choose 3.


Now we know from the white/black restriction that that takes out 6 possibilities. Therefore the answer is:

5c3-6= 20-6=14

what am i doing wrong?
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 [#permalink] New post 19 Feb 2007, 04:48
Tuneman wrote:
I would think that if they simply left out the restriction on the white and black pots, why wouldn't the answer be:5c3

you have 5 options, and you can choose 3.


Now we know from the white/black restriction that that takes out 6 possibilities. Therefore the answer is:

5c3-6= 20-6=14

what am i doing wrong?

It is not the just the selection you have to arrnage them as well. Also 5C3 is 10 and not 20.
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  [#permalink] 19 Feb 2007, 04:48
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