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08 Apr 2016, 03:13
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
61% (01:23) correct
39%
(01:29)
wrong
based on 272
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In how many ways can the letters of the word SPENCER be arranged if the S and P must always be together and the N and C must always be together?
A. 12
B. 24
C. 60
D. 120
E. 240
A. 12
B. 24
C. 60
D. 120
E. 240
17 Apr 2017, 15:59
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Bunuel
Take the task of arranging the letters and break it into stages.
Stage 1: Glue S and P together.
Note, this will ensure that S and P are together.
There are 2 ways to glue S and P together: SP and PS
So we can complete stage 1 in 2 ways
Stage 2: Glue N and C together.
There are 2 ways to glue N and C together: NC and CN
So we can complete stage 2 in 2 ways
IMPORTANT: We now have 5 "objects" to arrange. They are: E, E, R, S/P combo, N/C combo
Stage 3: arrange the 5 "objects" in a row
--------ASIDE-------------------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]
So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
---------BACK TO THE QUESTION--------------------
In stage 3, we must arrange 5 objects: E, E, R, S/P combo, N/C combo
There are 5 objects in total
There are 2 identical E's
So, the total number of possible arrangements = 5!/[(2!)] = 60
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus arrange all of the letters) in (2)(2)(60) ways (= 240 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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General Discussion
08 Apr 2016, 07:40
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There are 7 letters in the word SPENCER are 2E , SP , NC , R
We will consider SP and NC as a group and there are 2 E's
No of arrangements in which S and P must always be together and the N and C must always be together
= (5! *2 *2)/2!
= 5!*2
= 120 *2
= 240
Answer E
We will consider SP and NC as a group and there are 2 E's
No of arrangements in which S and P must always be together and the N and C must always be together
= (5! *2 *2)/2!
= 5!*2
= 120 *2
= 240
Answer E
