Vincen wrote: ↑

Wed Jun 24, 2020 1:53 am

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

[spoiler]OA=A[/spoiler]

Source: GMAT Prep

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

Vincen wrote: ↑

Wed Jun 24, 2020 1:53 am

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

[spoiler]OA=A[/spoiler]

Solution:

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