To furnish a room in model home, an interior decorator is to select 2

Question

To furnish a room in 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

A. 6
B. 8
C. 10 
D. 15
E. 30

The answer is A. BUT!
what is the next step in the process?

5C2*nC2=150
10*nC2=150
nC2=15

n! / 2!*(n-2)! = 15
n!=30(n-2)!

stuck here. what do I need to do next in order to arrive to the answer choice A, 6 tables?

award kudos for good explanation

Bunuel · Accepted Answer

You've done most of the work right, next step:
 \frac{n!}{2!*(n-2)!}=15 .  (n-2)!  will just cancel out and will get:

\frac{(n-1)*n}{2!}=15  -->  (n-1)*n=30  -->  n=6

Hope it's clear.

Bunuel · Answer

C(x,2)=\frac{x!}{2!*(x-2)!}=\frac{(x-2)!*(x-1)*x}{2!*(x-2)!}  so you can reduce by  (x-2)!  and you'll get  \frac{(x-1)x}{2} .

barakhaiev · Answer

Thank you very much, Bunuel, for such a fast and detailed response. Kudos, as promised!

shrouded1 · Answer

No of combinations = C(5,2) * C(x,2) = 10 * x(x-1)/2 =150
Where x is the number of tables

So x(x-1)/2 = 15
 x(x-1) = 30

And we know x is an integer

=> x=6

gettinit · Answer

Shrouded how did you arrive at x(x-1)/2 for c(x,2)? shouldn't this be x!/2!(x-2)! ? I am having trouble following how you arrived at x(x-1)/2? Please advise, thank you!

adishail · Answer

Out of curiosity

Why didn't we take different possible arrangements where order of selection is important (permutation). For example, if we have two chairs (A&B) and two tables (X&Y) selected from the warehouse, we can arrange them as AX & BY and AY & BX. Was this because of the fact that it was given that there are 150 different  combinations  ?

Bunuel · Answer

Decorator is to select 2 different chairs out of 5 and 2 different tables out of n. Now, two selections {C1, C2, T1, T2} and {C2, C1, T2, T1,} are the same for decorator as there are the same chairs and tables in both of them. So the order in which the items are taken is not considered.

Hope it's clear.

adishail · Answer

Yes. I understood your explanation.

However, I am talking about a case where we would have {C1T1, C2T2, and C1T2, C2T1} since these two can be considered different arrangements. But I think I somehow got stuck with the idea of making a set of a chair and table.

I guess if they would have worded it something like -  If 150 different sets of tables and chair are possible  ...........then order would have been important. Is that correct Bunuel ? And in that case we would have taken COMBINATION for chairs and PERMUTATION for tables or vice versa because if we would have taken PERMUTATION for both tables an chairs, then we would have ended up with two identical cases like {C1T1, C2T2} and {C2T2, C1T1}.

Bunuel · Answer

You are right we are not told to pair a chair with table. Also I don't think that changing the word "combination" with "set" would make a difference, the question is clearly talking about the different selections of 2 chairs and 2 tables.

If we were interested in such pairs then you should notice that one particular selection of 2 chairs and 2 tables {C1, C2, T1, T2} can give 2 different pairs of chair/table: {C1/T1, C2/T2} or {C1/T2, C2/T1}.

cabelk · Answer

Combo box arrangement for chairs
(_)(_)/2!

Combo box arrangement for chairs and tables
(_)(_)/2! * (_)(_)/2!

Fill in combo box arrangement
(5)(4)/2! * (_)(_)/2!

(5)(4)/2! * (_)(_)/2! = 150

(5)(4)/2! * (Tables)(Tables - 1)/2! = 150

Algebra
(Tables)(Tables - 1) = 150 * 2! * 2 ! / (5)(4) = 30 = 6*5
Tables = 6

Bunuel · Answer

To furnish a room in 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

C^2_5*C^2_t=150 , where  t  is the # of tables -->  C^2_5*C^2_t=150  -->  C^2_t=15  -->  \frac{t!}{2!(t-2)!}=15  -->  (t-2)!  will cancel out -->  (t-1)t=30  -->  t=6 .

Answer: A.

vyassaptarashi · Answer

GMAT prep problem... 
It should be solved like following 
Suppose we have to select 2 tables from “X” tables
We have already given that we have to select 2 chairs form 5 and we total possible combinations are 150

Therefore, 
5C2 * XC2 = 150
5*4/2 * X!/2!(X-2)! = 150
X!/2!(X-2)! = 15
X!/(X-2)! = 30
X(X-1)(X-2)! / (X-2)! = 30 
X(X-1) = 30
X2 – X – 30 = 0 
X2 – 6X + 5X – 30 = 0
X(X – 6) + 5(X – 6) = 0
(X – 6) (X + 5) = 0
X = 6 or X = -5
Negative value is not possible since we are calculating the number of tables
Therefore X = 6
Hence “A” is the answer

JeffTargetTestPrep · Answer

We are given that an interior decorator is to select 2 chairs and 2 tables from a collection of chairs and tables. We are also given that there are 5 chairs in the warehouse and 150 different possible combinations. We must determine the number of tables. We can let n = the number of tables and create the following equation:
 
5C2 x nC2 = 150
 
  x   = 150
 
20/2 x (n^2 – 1)/2 = 150
 
10 x (n^2 – 1)/2 = 150
 
(n^2 – 1)/2 = 15
 
n^2 – 1 = 30
 
n^2 – 1 – 30 = 0
 
(n – 6)(n + 5) = 0
 
n = 6 or n = -5.
 
Since n must be positive, the number of tables is 6.
 
Answer: A

IloveMBA123 · Answer

Hi sir, can you show us the detail worksteps how about to cancel out from (N-2)! to (n-1)^N?

Many thanks

Bunuel · Answer

Please read the whole thread before posting a question: https://gmatclub.com/forum/to-furnish-a ... ml#p789229

BrentGMATPrepNow · Answer

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,  N C2 = 15
Our goal is to find the value of N.

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

Answer = A

Cheers,
Brent

NathanMasky · Answer

1. Calculate the number of ways to choose chairs:

The interior decorator needs to choose 2 chairs out of 5. This is a combination problem (order doesn't matter), and the formula for combinations is:

nCr = n! / (r! * (n-r)!)

Where n is the total number of items, and r is the number you choose.

So, the number of ways to choose 2 chairs from 5 is:

5C2 = 5! / (2! * 3!) = (5 * 4 * 3 * 2 * 1) / ((2 * 1) * (3 * 2 * 1)) = 10

2. Set up an equation:

Let 't' be the number of tables in the warehouse. The decorator needs to choose 2 tables from 't' tables. The number of ways to do this is tC2.

We know that the total number of combinations of chairs and tables is 150. Since the choices of chairs and tables are independent, we multiply the number of ways to choose chairs by the number of ways to choose tables to get the total number of combinations:

(Number of ways to choose chairs) * (Number of ways to choose tables) = Total combinations

10 * tC2 = 150

3. Solve for 't':

First, divide both sides of the equation by 10:

tC2 = 15

Now, use the combination formula for tC2:

t! / (2! * (t-2)!) = 15

Simplify:

(t * (t-1)) / 2 = 15

Multiply both sides by 2:

t * (t-1) = 30

Expand:

t^2 - t = 30

Rearrange to form a quadratic equation:

t^2 - t - 30 = 0

Factor the quadratic:

(t - 6)(t + 5) = 0

The possible solutions for 't' are 6 and -5. Since the number of tables cannot be negative, the number of tables is 6.

Answer: There are 6 tables in the warehouse.

To furnish a room in model home, an interior decorator is to select 2

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