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
To furnish a room in model home, an interior decorator is to
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Since 150 combinations are possible, we get: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
(number of ways to choose 2 chairs)(number of ways to choose 2 tables) = 150.
From 5 chairs, the number of ways to choose 2 = 5C2 = (5*4)/(2*1) = 10.
Thus:
(10)(number of ways to choose 2 tables) = 150
number of ways to choose 2 tables = 150/10 = 15.
We can PLUG IN THE ANSWERS -- which represent the number of tables -- to see which yields 15 ways to choose 2 tables.
B: 8
From 8 tables, the number of ways to choose 2 = 8C2 = (8*7)/(2*1) = 28.
Too big.
A smaller answer choice is needed.
The correct answer is A.
A: 6
From 6 tables, the number of ways to choose 2 = 6C2 = (6*5)/(2*1) = 15.
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Total # of combinations = (# of ways to select 2 chairs)(# of ways to select 2 tables)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
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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$$? = T$$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
$$C\left( {T,2} \right) \cdot \underbrace {C\left( {5,2} \right)}_{ = \,\,10} = 150\,\,\,\,\, \Rightarrow \,\,\,\,\,C\left( {T,2} \right) = 15$$
$${{T\left( {T - 1} \right)} \over 2} = 15\,\,\,\,\,\,\mathop \Rightarrow \limits^{T\, > \,0} \,\,\,\,\,\,? = T = 6$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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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: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
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
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