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Advanced Combinatorics Problem [#permalink]
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?

Re: Advanced Combinatorics Problem [#permalink]
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?

Re: Advanced Combinatorics Problem [#permalink]
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

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

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 _________________

"Live as if you were to die tomorrow. Learn as if you were to live forever." - Mahatma Gandhi

ANSWER is explained [#permalink]
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

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? _________________

"Live as if you were to die tomorrow. Learn as if you were to live forever." - Mahatma Gandhi

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.