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In how many ways can the letters of the word SPENCER be...

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In how many ways can the letters of the word SPENCER be...

by swerve » Tue Mar 20, 2018 2:23 pm
In how many ways can the letters of the word SPENCE be arranged if the S and P must always be together and N and C must always be together?

A. 12
B. 24
C. 60
D. 120
E. 240

The OA is E.

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.
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by regor60 » Wed Mar 21, 2018 5:40 am
Assuming you mean the word SPENCER as in the title vs the text of your question.

There are 7 spaces for each letter, but since S and P must be one unit and N and C also must be one unit, there are 5 spaces to arrange the letters.

So there are 5! permutations.

However, within the SP and NC pairings, there are two ways to order the letters, so there are 5!x2x2 possible arrangements.

Finally, because there are two E's and the above gives credit twice for a pair of E's, the number of distinct permutations is

5!x2x2/2 = 240, E
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by Vincen » Thu Mar 22, 2018 2:55 am
Hello.

Since SPENCER has 7 letters the number of ways to arrange the letters is 7!

Now, since there are two E's, we included equal cases before, for example: SPENCeR and SPeNCER. Both words are equal. To avoid repeating cases, we have to divide by 2. Hence, we have 7! / 2.

Now, we are said that the letters S and P must always be together, hence we consider SP as one single letter. Therefore, the number of ways to arrange now is 6!

Also, we are said that the letters N and C must always be together, hence we consider NC as one single letter. Thus, the number of ways to arrange the letters now is 5!

Finally, since we can write SP or PS, and NC or CN, we need to multiply by 2 twice. In conclusion, the number of ways to arrange the letters with the given condition is: $$\frac{5!}{2}\cdot2\cdot2\ =5!\cdot2\ =\ 120\cdot2=240.$$ Therefore, the answer is the option E.
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by GMATGuruNY » Thu Mar 22, 2018 3:16 am
The posted prompt has a typo: the word SPENCE should actually written as SPENCER, as follows:
swerve wrote:In how many ways can the letters of the word SPENCER be arranged if the S and P must always be together and N and C must always be together?

A. 12
B. 24
C. 60
D. 120
E. 240
Since S and P must be together and N and C must be together, consider each pair a SEPARATE BLOCK in the arrangement, yielding the following list of 5 elements:
[SP], [NC], E, E, R
The number of ways to arrange 5 elements = 5! = 120.

Since SP can be reversed to PS -- doubling the number of arrangements -- we multiply by 2:
120*2 = 240.
Since NC can be reversed to CN -- again doubling the number of arrangements -- we again multiply by 2:
240*2 = 480.

The arrangement includes 2 identical E's.
When an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical E's:
480/2! = 240.

The correct answer is E.
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by deloitte247 » Fri Mar 23, 2018 11:13 pm
Since [S and P] and [N and C] must always be together and we will arrange the element thus:
$$\left[SP\right],\left[NC\right],E,E=4\ objects$$
Note: if there are 'n' objects where 'A' of them are alike <identical> and <B> of them are also alike <identical>...; total numbers of possible arrangements =m $$\ \frac{n!}{\left[<A!><B!>...\right]}\ in\ \left[SP\right],\ \left[NC\right],\ E,\ E$$
There are 4 objects
There are 2 identical E's. [SP] and [NC] are 2 objects with 2 possibilities= $$\frac{\left(\ 4!\left(2\right)\left(2\right)\right)}{\left(2!\right)}=\ \frac{\left[4\cdot3\cdot2\cdot1\cdot\left(2\right)\left(2\right)\right]}{2\cdot1}$$
$$=\frac{\left[24\left(2\right)\left(2\right)\right]}{2}=48\ ways$$
I found 48 as the accurate answer which is not one of the option
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