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Furnished Combinatorics

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Furnished Combinatorics

by szDave » Thu Jan 24, 2013 6:26 am
Hello,

To furnish a room in a model home, an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tables in a warehouse that are all different from each other. If there are 5 chairs in the warehouse and if 150 different combinations are possible, how many tables are in the warehouse?

a) 6
b) 8
c) 10
d) 15
e) 30

I started computing with the anagram method and after a point, didn't know how to proceed. here is my calculation:

5!/(2!*3!) = 10 - so there are 10 combinations for the chairs. That means that there are 150 - 10 combinations for the tables. That yields 140 = x!/[2!*(x-2)!]. But I didn't know how to solve this, so picked the answer choices and none yielded A.

How do you get the right answer?

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by [email protected] » Thu Jan 24, 2013 6:44 am
szDave wrote:Hello,

To furnish a room in a model home, an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tables in a warehouse that are all different from each other. If there are 5 chairs in the warehouse and if 150 different combinations are possible, how many tables are in the warehouse?
a) 6
b) 8
c) 10
d) 15
e) 30
Total # of combinations = (# of ways to select 2 chairs)(# of ways to select 2 tables)
So, 150 = (# of ways to select 2 chairs)(# of ways to select 2 tables)

# of ways to select 2 chairs
5 tables, choose 2 of them.
Since the order of the selected chairs does not matter, we can use combinations.
This can be accomplished in 5C2 ways (10 ways)

Total # of combinations = (# of ways to select 2 chairs)(# of ways to select 2 tables)
150 = (10)(# of ways to select 2 tables)
(# of ways to select 2 tables) = 15

# of ways to select 2 tables
Let N = # of tables.
We have N tables, choose 2.
This can be accomplished in NC2 ways
So, NC2 = 15
Our goal is to find the value of N.

From here, we can just start checking answer choices.
We get 6C2 = 15, so N = 6, which means there are 6 tables.

If anyone is interested, we have a free video on calculating combinations (like 6C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Cheers,
Brent
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by gmattesttaker2 » Mon Feb 04, 2013 6:58 pm
[email protected] wrote:
szDave wrote:Hello,

To furnish a room in a model home, an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tables in a warehouse that are all different from each other. If there are 5 chairs in the warehouse and if 150 different combinations are possible, how many tables are in the warehouse?
a) 6
b) 8
c) 10
d) 15
e) 30
Total # of combinations = (# of ways to select 2 chairs)(# of ways to select 2 tables)
So, 150 = (# of ways to select 2 chairs)(# of ways to select 2 tables)

# of ways to select 2 chairs
5 tables, choose 2 of them.
Since the order of the selected chairs does not matter, we can use combinations.
This can be accomplished in 5C2 ways (10 ways)

Total # of combinations = (# of ways to select 2 chairs)(# of ways to select 2 tables)
150 = (10)(# of ways to select 2 tables)
(# of ways to select 2 tables) = 15

# of ways to select 2 tables
Let N = # of tables.
We have N tables, choose 2.
This can be accomplished in NC2 ways
So, NC2 = 15
Our goal is to find the value of N.

From here, we can just start checking answer choices.
We get 6C2 = 15, so N = 6, which means there are 6 tables.

If anyone is interested, we have a free video on calculating combinations (like 6C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Cheers,
Brent

Hello Brent,

I was trying to use the counting method here:

1 possible way to select 2 chairs and 2 tables is:

5 * 4 * t * (t-1)
----- --------
2! 2!

So here the order of selection is : C C T T

Another possibility is C T C T

So total possibilites are 4 * 3 * 2 * 1 = 24

5 * 4 * t * (t-1)
----- -------- * 24 = 150
2! 2!

However, I am not sure if this is correct since I am getting something like:

4t^2 - 4t - 5 = 0

I was wondering if you can please assist here? Thanks a lot for your valuable time and help.

Best Regards,
Sri

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by [email protected] » Tue Feb 05, 2013 8:37 am
gmattesttaker2 wrote:
Hello Brent,

I was trying to use the counting method here:

1 possible way to select 2 chairs and 2 tables is:

5 * 4 * t * (t-1)
----- --------
2! 2!

So here the order of selection is : C C T T

Another possibility is C T C T

So total possibilites are 4 * 3 * 2 * 1 = 24

5 * 4 * t * (t-1)
----- -------- * 24 = 150
2! 2!

However, I am not sure if this is correct since I am getting something like:

4t^2 - 4t - 5 = 0

I was wondering if you can please assist here? Thanks a lot for your valuable time and help.

Best Regards,
Sri
Hey Sri,

You were almost there.
When you got to . . .

5 * 4 * t * (t-1)
----- --------
2! 2!

. . . your expression was perfect. The 2! in the denominator of each fraction already accounted for certain counting arrangements more than once.

For example,
(5)(4)/2! represents the number of ways we can select 2 chairs from 5 chairs.
Select 1st chair: accomplished in 5 ways (5 chairs to choose from)
Select 2nd chair: accomplished in 4 ways (4 chairs to choose from)
(5)(4)= 20, however we've counted each selection twice.
Example: selecting A then B is the same as selecting B then A, however our above calculation treats these as separate selections.
So, to deal with this duplications, we divide (5)(4) by 2! (the number of ways to arrange 2 items).
Perfect!

So, in both cases, you correctly divided the number of arrangements by 2!.
So, that extra calculation (with 4!) was unnecessary)

This means we're ready for an equation:
(5)(4)(t)(t-1)/(2!)(2!) = 150
Simplify to get: (t)(t-1) = 30
Here, we can see that t = 6

Cheers,
Brent
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by Lifetron » Fri Feb 08, 2013 2:09 am
5c2 * xc2 = 150
xc2 = 15
x*(x-1)/2 = 15
x(x-1) = 30
so, x = 6

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by gmattesttaker2 » Sun Mar 03, 2013 8:25 pm
[email protected] wrote:
szDave wrote:Hello,

To furnish a room in a model home, an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tables in a warehouse that are all different from each other. If there are 5 chairs in the warehouse and if 150 different combinations are possible, how many tables are in the warehouse?
a) 6
b) 8
c) 10
d) 15
e) 30
Total # of combinations = (# of ways to select 2 chairs)(# of ways to select 2 tables)
So, 150 = (# of ways to select 2 chairs)(# of ways to select 2 tables)

# of ways to select 2 chairs
5 tables, choose 2 of them.
Since the order of the selected chairs does not matter, we can use combinations.
This can be accomplished in 5C2 ways (10 ways)

Total # of combinations = (# of ways to select 2 chairs)(# of ways to select 2 tables)
150 = (10)(# of ways to select 2 tables)
(# of ways to select 2 tables) = 15

# of ways to select 2 tables
Let N = # of tables.
We have N tables, choose 2.
This can be accomplished in NC2 ways
So, NC2 = 15
Our goal is to find the value of N.

From here, we can just start checking answer choices.
We get 6C2 = 15, so N = 6, which means there are 6 tables.

If anyone is interested, we have a free video on calculating combinations (like 6C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Cheers,
Brent
Hello Brent,

Hope all is well. I was just wondering if the following approach would be correct here:

Probability of 1 possible outcome is:
(2/5)(1/4)(2/t)(2/(t-1))

= 1/5(t^2 - t)

I was then trying to find the total number of possible outcomes using the MISSISSIPPI approach like this problem:

https://www.beatthegmat.com/probability- ... tml#481625

So I was getting something like:

(4 x 3 x 2 x 1)/(2! x 2!)

Basically I am trying to get all the possible combinations of 2 chairs and 2 tables

i.e. C1 C2 T1 T2
C1 T1 C2 T1

etc. and then dividing by 2!x 2! to eliminate duplicates.

So I now have:

(6) x ( 1/(5t^2 - 5t)) = 150

However, I am not sure if this approach is correct here? Can you please help? Thank you very much for all your valuable time and help.

Best Regards,
Sri

GMAT/MBA Expert

GMAT Instructor
Posts: 15056
Joined: 08 Dec 2008
Location: Vancouver, BC
Thanked: 5254 times
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by [email protected] » Mon Mar 04, 2013 8:33 am
gmattesttaker2 wrote: Hello Brent,

Hope all is well. I was just wondering if the following approach would be correct here:

Probability of 1 possible outcome is:
(2/5)(1/4)(2/t)(2/(t-1))

= 1/5(t^2 - t)

I was then trying to find the total number of possible outcomes using the MISSISSIPPI approach like this problem:

https://www.beatthegmat.com/probability- ... tml#481625

So I was getting something like:

(4 x 3 x 2 x 1)/(2! x 2!)

Basically I am trying to get all the possible combinations of 2 chairs and 2 tables

i.e. C1 C2 T1 T2
C1 T1 C2 T1

etc. and then dividing by 2!x 2! to eliminate duplicates.

So I now have:

(6) x ( 1/(5t^2 - 5t)) = 150

However, I am not sure if this approach is correct here? Can you please help? Thank you very much for all your valuable time and help.

Best Regards,
Sri
Hi Sri,

I'm not sure why you have injected probability into this question. This is a counting question.
Your earlier solution in this thread (with my edits) was a good approach.

Cheers,
Brent
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