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A,A,B,B,B,C,D,E arrangement question

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A,A,B,B,B,C,D,E arrangement question

by ithamarsorek » Fri Dec 24, 2010 12:51 pm
Does anyone have a simpler explanation? I got most of it but trying to really understand expecially the part of dividing everything by 2 towards the end.
Thanks much and happy holiday!


In how many different ways can the letters A, A, B, B, B, C, D, E be arranged if the letter C must be to the right of the letter D?

1,680
2,160
2,520
3,240
3,360


Analyze the Question:

If you see a group and you are asked to arrange items in the group, you should immediately expect to calculate permutations or combinations.

Identify the Task:

Remember that with permutations order matters. If order does not matter, then we are dealing with combinations. We must use these formulas to find the number of possible orderings of the given letters, remembering to account for the "C is to the right of D" requirement.

Approach Strategically:

We have a total of 8 letters. Some of the letters appear more than once. The requirement of the question stem that the letter C is to be to the right of the letter D, with the letters D and C each appearing once.

If n is a positive integer, the number of different ways to arrange n different objects is n!, where n! is the product of the first n positive integers. Thus, n! = n (n − 1)(n − 2) ... (3)(2)(1).

If we had 8 different letters, the number different of ways to arrange the 8 letters would be 8! = (8)(7)(6)(5)(4)(3)(2)(1). However, this number will be lower because of duplicates in this problem. Because the letter A appears twice, we must divide 8! by 2! Because the letter B appears 3 times, we must divide 8! by 3!. So the number of different ways to arrange the letters A, A, B, B, B, C, D, E is . We also have the requirement that the letter C is to the right of the letter D. For any specified locations of the letter C and the letter D, there are the same number of arrangements for the other 6 letters when C is to the right of D and D is to the right of C. So we must divide the number of possible arrangements of all the letters by 2. So the number of different ways to arrange the letters A, A, B, B, B, C, D, E where the letter D is to the right of the letter C is , which is . To find the answer to the question, let's find the value of . We have = = = (4)(7)(3)(5)(4) = 1,680. Answer Choice (A) is correct.

Confirm your Answer:

Remember if you are canceling out factorials in a fraction, you need to clearly mark what is left

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by anshumishra » Fri Dec 24, 2010 1:01 pm
A -> 2 times
B - >3 times
C,D,E -> 1 time

Total number of arrangements = 8!/(2!*3!)
Now out of these arrangements in half of them C will appear right and in half of them to left, so the number of arrangements = 8!(2!*3!)*2 = 1680
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by ithamarsorek » Fri Dec 24, 2010 1:40 pm
How do you know where it appears half of the time ?

Is it like a coin toss when we know that the chance for heads is 1/2 ans tails is 1/2 ?

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by anshumishra » Fri Dec 24, 2010 1:48 pm
ithamarsorek wrote:How do you know where it appears half of the time ?

Is it like a coin toss when we know that the chance for heads is 1/2 ans tails is 1/2 ?
Since there is not any weird rules added (such as C should be in corner, blah, blah...), the probability of C being on left side of D would be equal to being it on right side. Also, since "E" appears just once like C and D, so if you are interested to know the probability of E being one side of C or D would be equal to E being the other side of C or D.

To make sure that it is not just an hypothesis, I would advice you to play with these cases (Extend it if you want) :
A,B,C,D - > What is the probability of B being left side of C ?
A,A,B,C,D -> What is the probability of B being left side of C ?
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by ithamarsorek » Fri Dec 24, 2010 2:06 pm
ABCD B on the left of C?

4! = 24
24/2 = 12

AABCD B on left of C?

5!/2! = 5*4*3 = 60

60/2 = 30

?

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by anshumishra » Fri Dec 24, 2010 2:13 pm
ithamarsorek wrote:ABCD B on the left of C?

4! = 24
24/2 = 12

AABCD B on left of C?

5!/2! = 5*4*3 = 60

60/2 = 30

?
The answer looks good to me. You know now, how to use this fact to your advantage to save some time ?
When you get some time, try to solve these questions with other approaches and then compare with this method; then you would be more confident to use these kind symmetrical scenarios.

Hope that helps !
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Anshu

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by beat_gmat_09 » Fri Dec 24, 2010 9:40 pm
Thought i'd share this-
Consider 2 examples , first with 3 persons, second with 6.
Take for example there are 3 persons EMX, and E should be behind M. Let's see how can this be arranged.

E _ _
_ E _


Now 1st case - M can occupy 2 places E M X or E X M, hence number of ways = 2
2nd case - M can occupy only one place X E M number of ways = 1
Total ways = 2+1 = 3 = 3!/2 = 3

Now take 6 contestants (just to understand how the answer is coming n!/2)

E _ _ _ _ _
_ E _ _ _ _
_ _ E _ _ _
_ _ _ E _ _
_ _ _ _ E _

1st - M can occupy 5 places, once M is fixed there are other 4 places which can arrange in 4! Ways, thus number of places in 1st case = 5*4!=120

2nd - M can occupy 4 places, once M is fixed there are 4 places which can be arranged in 4! Ways, total number of ways = 4*4! = 96

3rd - similarly 3*4! = 72

4th - similarly 2*4! = 48

5th - similarly 1*4! = 24

Add all the cases-
Total number of ways = 4!(5+4+3+2+1) = 4!*15 = 360 ways
Which is same as 6!/2
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