Bunuel wrote:
Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be seated in a single row of six chairs. If Betsy cannot sit next to Emily, how many different arrangements of the six children are possible?
A. 240
B. 480
C. 540
D. 720
E. 840
One more approach:
Take the task of arranging the 6 students in a row and break it into
stages.
NOTE: We're going to
ignore the chairs and just examine how many ways there are to arrange the 6 children in a row.
The answer will be the same.
Stage 1: Arrange Arya, Chen, Daniel and Franco is a row
We can arrange n unique "objects" in a row in n! ways
So, we can arrange these 4 children in 4! ways (24 ways)
So, we can complete stage 1 in
24 ways
IMPORTANT STEP: For each of the 24 arrangements of Arya, Chen, Daniel and Franco,
add a space on either side of each child.
For example, one of the possible arrangements is Chen, Daniel, Franco, Arya
So, add spaces as follows: ___Chen___Daniel___Franco___Arya___
We'll now place Betsy in a space and place Emily in a different space.
This will ENSURE that Betsy and Emily are not seated together.
Stage 2: Choose a space to place Betsy
There are 5 spaces to choose from.
So we can complete stage 2 in
5 ways
Stage 3: Choose a space to place Emily
Once we select a space in stage 2, there are 4 spaces remaining to choose from.
So we can complete stage 3 in
4 ways
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus arrange all 6 children) in
(24)(5)(4) ways (= 480 ways)
Answer:
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.
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I used another approach you teach. Seems easier.
Started with the most restrict: (1* x 4 x 4 x 3 x 2 x 1) = 96 possibilities. *means Betsy or Emily. Then, multiplied by 5 different chairs they could be seated 96 x 5 = 480.