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

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

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Source: GMAT Prep

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

The OA is A.

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BTGmoderatorLU wrote:
Source: GMAT Prep

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

The OA is A.
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.

Answer = A

Cheers,
Brent

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BTGmoderatorLU wrote:
Source: GMAT Prep

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
\[? = T\,\,\,\left( {\# \,\,{\text{tables}}} \right)\]
\[10 \cdot C\left( {T,2} \right) = C\left( {5,2} \right) \cdot C\left( {T,2} \right) = 150\,\,\,\,\,\, \Rightarrow \,\,\,\,\,C\left( {T,2} \right) = 15\]
\[C\left( {T,2} \right) = 15\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\left\{ \begin{gathered}
T\left( {T - 1} \right) = 30 \hfill \\
6 \cdot 5 = 30\,\,\,\,\left( {**} \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = T = 6\]
\[\left( * \right)\,\,\,C\left( {T,2} \right) = \underleftrightarrow {\frac{{T\left( {T - 1} \right)\left( {T - 2} \right)!}}{{2!\,\,\left( {T - 2} \right)!\,\,}}} = \frac{{T\left( {T - 1} \right)}}{2}\]
\[\left( {**} \right)\,\,\,\left( { - 5} \right) \cdot \left( { - 6} \right) = 30\,\,\,\,\, \Rightarrow \,\,\,\,T = - 5\,\,\,\,\left( {{\text{not}}\,\,{\text{viable}}} \right)\]
(In words: we have a second-degree equation in the variable T, and we found the two real distinct roots of it: 6 and -5. From the fact that -5 is negative, our answer is really 6.)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

_________________
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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BTGmoderatorLU wrote:
Source: GMAT Prep

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
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

[(5 x 4)/2!] x [(n x (n - 1))/2!] = 150

20/2 x (n^2 - n)/2 = 150

10 x (n^2 - n)/2 = 150

(n^2 - n)/2 = 15

n^2 - n = 30

n^2 - n - 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

_________________
Scott Woodbury-Stewart Founder and CEO

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