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Slight Variation of a different problem

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Slight Variation of a different problem

by knight247 » Fri Sep 30, 2011 10:55 pm
In how many different ways can the letters A,A,B,B,B,C,D be arranged if the letter C must be to the right of the letter D?

Considering there are an odd number of elements, Will C still be to the right of D in half of all the arrangements with no restrictions and D to the right of C in the other half???
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by phoenix111 » Fri Sep 30, 2011 11:14 pm
1st part: Possible arrangement of AABBB : 5!/2!3! = 120/12 = 10
Now we insert C and D in these arrangements :

_X_X_X_X_X_

if C in 1st '_' : possible places for D : 5
Similarly for other positions of C possible values : 4,3,2,1
Total : 15

Total possible arrangements : 10*15 = 150


2nd part:

Yes.
If we interchange positions positions of C & D in the above questions,
we will again get 150 new possible arrangements.

Now just to verify, total arrangments without restrictions :

Out of 6 '_' we select two 2 : 6C2
Now they can either have CD or DC.
So total : 6p2 = 30
Total arrangements : 10*30 = 300
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by GmatMathPro » Fri Sep 30, 2011 11:20 pm
Having an odd number of elements wouldn't affect that principle. Picture every single unique arrangement of these terms written out. Then imagine you went through and lined up all of the ones with C to the left of D in one column. For each one of those arrangements, we could find a different arrangement where the positions of C and D are swapped, with D to the left of C. It could look something like this:

AABBBCD AABBBDC
ABCBBDA ABDBBCA
CADABBB DACABBB

and so on. We know that every single sequence in the left column will have a mate that we can put in the right column, so it must be half. You might try experimenting with a smaller number of letters that allows you to actually write all of the sequences out to verify that it's true, like ABC, with B to the left of C:

ABC ACB
BAC CAB
BCA CBA

So, 7!/(2!3!)=420 ways to arrange the letters, and 210 with C to the right of D.
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