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Forbidden arrangement question

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Forbidden arrangement question

by k4lnamja » Mon Dec 20, 2010 10:38 pm
Hello,

I feel a little silly asking this but I'm blaming it on it being late. Anyways, I've a question about forbidden arrangements and I'm hoping someone could show me how they got the answer.

"in how many different ways can the letters in the word REVIEW be arranged if the two E's cannot be adjacent"? Correct answer is 240 but I keep getting 360.



Also, "there are 2 teachers and 3 students that need to stand in a line. how many times can you arrange the teachers so the teachers are arranged at the end?"

The answer is 12 but I dont know why either.

I appreciate it. Thank you
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by diebeatsthegmat » Tue Dec 21, 2010 4:57 pm
k4lnamja wrote:Hello,

I feel a little silly asking this but I'm blaming it on it being late. Anyways, I've a question about forbidden arrangements and I'm hoping someone could show me how they got the answer.

"in how many different ways can the letters in the word REVIEW be arranged if the two E's cannot be adjacent"? Correct answer is 240 but I keep getting 360.



Also, "there are 2 teachers and 3 students that need to stand in a line. how many times can you arrange the teachers so the teachers are arranged at the end?"

The answer is 12 but I dont know why either.

I appreciate it. Thank you
2 teachers at the end so there is 2! ways to arrange em
3 people left = 3! ways
total=2!*3!=12
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by beat_gmat_09 » Tue Dec 21, 2010 8:16 pm
Hi,
The word - REVIEW contains 6 alphabets with 2 E's
Tie these E's together - RVIWEE consider EE as 1 entity total entities = 5, number of ways to arrange these = 5! = 120
The order of E's doesn't matter here as they are same - indistinguishable.
Total number of ways to arrange 6 alphabets with 2 E's (again the order of E does not matter - using the p!/q!*r!.. rule) = 6!/2! = 720/2 = 360
The number of ways two E's are not adjacent is = Total number of ways of arranging - Total number of ways when E's are adjacent.

Total number of ways two E's cant be adjacent = 360 - 120= 240 ways.
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by GMATGuruNY » Tue Dec 21, 2010 8:24 pm
k4lnamja wrote:.

"in how many different ways can the letters in the word REVIEW be arranged if the two E's cannot be adjacent"? Correct answer is 240 but I keep getting 360.
Good arrangements = Total possible arrangements - Bad arrangements

Total possible ways to arrange REVIEW = 6!/2! = 360. (Number of ways to arrange 6 elements is 6!. We divide by 2! to account for the two EEs, which will reduce the number of unique arrangements.)

A bad arrangement places the two EE's adjacent to each other. We need to count the number of ways we can arrange EE and the other 4 letters. Since we're now arranging 5 elements (EE,R,V,I,W), the number of bad arrangements = 5! = 120.

Thus, good arrangements = 360-120 = 240.
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by Night reader » Tue Dec 21, 2010 9:28 pm
k4lnamja wrote:Hello,

I feel a little silly asking this but I'm blaming it on it being late. Anyways, I've a question about forbidden arrangements and I'm hoping someone could show me how they got the answer.

"in how many different ways can the letters in the word REVIEW be arranged if the two E's cannot be adjacent"? Correct answer is 240 but I keep getting 360.



Also, "there are 2 teachers and 3 students that need to stand in a line. how many times can you arrange the teachers so the teachers are arranged at the end?"

The answer is 12 but I dont know why either.

I appreciate it. Thank you
REVIEW is a 6 letter word. Normally, 6 letter words could be arranged in 6! ways. However, in this case the letter E appears twice. Therefore the number of different arrangements that are possible 6!/2!

The number of arrangements in which the two Es occur together can be found by construing both the Es to be joined together as a single letter. Let us say we represent EE as Q. Now in the word REVIEW, we have the two Es joined together and have it replaced with Q. Therefore, there will be 5 letters in the word REVIEW or Q(instead of EE), R, V, I, W [please note, if you had more repeating letters you would need to substitute single letters for all of them at once]. This can be arranged as 5!

What we require is the number of arrangements where the two Es do not occur together
= (Total number of arrangements) - (Number of arrangements where the two Es occur together)

= 6!/2! - 5!

there are in total of 5 people - 2 teachers and 3 students, that need to stand in a line. We have restriction that the teachers stand (are arranged) at the end. So start with restrictions:

__ __ __ __ __ five slots or places - five people may stand

1st 2nd 3rd 4th 5th
1st 2nd 3rd -- 2 teachers may stand in place No 4 or 5-- 1 teacher may stand in place No 5 =>
__ __ __ 2 * 1

--three students may stand in place No 1-- two students may stand in place No 2-- one student may stand in place No 3 and -- 2 teachers may stand in place No 4 or 5-- 1 teacher may stand in place No 5

3 * 2 * 1 * 2 * 1

total 12 ways


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