please help explain ... permutation/comb prob??

karenmeow

I copy and pasted from a Kaplan online test (some of the parts are missing in the explanation). In the explanation, it says to divide by 2 because C has to be to the right of D. Where does the 2 come from?? I am lost here. Thanks!

Q: 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

A: 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). 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 8!/(2!3!). 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 1/2 (8!/(2!3!)), which is 8!/((2)2!3!). To find the answer to the question, let's find the value. Choice (A) is correct.

pranavc · Posted: Mon Jul 07, 2008 1:25 pm Post subject:

I guess I've gotten quite rusty with my permutations/combinations concepts, but why exactly you divide the total number of possibilities by 2! and 3! for the A's and B's respectively? It would be great if you could fill me in on this bit. Thanks in advance.

evansbd · Posted: Thu Jul 17, 2008 8:01 am Post subject: Insanity

This question is ridiculous...!

sudhir3127 · Posted: Thu Jul 17, 2008 8:31 am Post subject:

i think the 2 is because number of way C and D can arrange is 2! which is 2.

CITI29 · Posted: Thu Jul 17, 2008 12:54 pm Post subject:

but if condition is that 'c' has to be to the right of 'd'...then dont we need to treat them as 'fixed' alphabets?.

In such a case solution shld be 7!/ (2!3!).

Pls let me know if my understaning is wrong.

preetha_85 · Posted: Thu Jul 17, 2008 8:24 pm Post subject:

Hi
The solution goes like this :
U have 8 alphabets nd the condition given is C is to the right of D.
Assume D is in the first pos. then C can g o into any of the other 7 slots nd so can any other alphabet given(i.e 7! ways). But since B is repeated 3 times nd A is repeated 2 twice.. this combination now becomes (7!/(3!*2!)
Now if D is in the 2 nd slot then C can go into only 6 slots (coz of the condition given) then the other alphabets can be arranged in any of the other 6 slots.. so this combination will be (6*(6!/(3!*2!))
Now if D is in 3rd slot then .. (5*(6!/(3!*2!))
nd so on till D is in the last before slot 1*( 6!/(3!*2!))
D in the last slot is not possible.

So the answe will be the sum of all these quantities i.e :
(7!/(3!*2!) + 6!/(3!*2!)) (6+5+4+3+2+1)
i.e 1680

spanlength · Posted: Fri Jul 18, 2008 1:58 am Post subject:

We divide by 2 because C can be either right of D or left of D.So in half of the combination its to the left other half it will to the right.

moliver · Posted: Fri Jul 18, 2008 4:45 am Post subject:

CITI29 · Posted: Fri Jul 18, 2008 1:19 pm Post subject:

ahhhhhh....now I get it...thanks guys!