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?
A) 1680
B)2160
C) 2520
D)3240
E) 3360
A) 1680
B)2160
C) 2520
D)3240
E) 3360
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
Can you please explain the above in detail. Appreciate.dmateer25 wrote:This one is actually easier than it seems.
So you have the first part: 8!/2!3! = 3360
This is the total number of arrangements.
Now, in the arrangement of the 8 letters, C will be to the right of D in exactly half of the cases. Therefore, you need to divide 3360 by 2.
3360/2 = 1680
A
Epitome: Do not permute (DC).[email protected] wrote:Can you please explain the above in detail. Appreciate.dmateer25 wrote:This one is actually easier than it seems.
So you have the first part: 8!/2!3! = 3360
This is the total number of arrangements.
Now, in the arrangement of the 8 letters, C will be to the right of D in exactly half of the cases. Therefore, you need to divide 3360 by 2.
3360/2 = 1680
A
It will be better to break the problem by omitting the repeating letters.gmatassistance wrote:I'm having trouble determining how to calculate for the limitation (C must be right of D) . . .
I understand that there are 8 numbers with two 'limitations' (i.e. repeat letters A and B) hence the total possible solutions could be 8! / 2! x 3!
Can someone please explain in detailed steps? Thanks in advance!
This is a very important concept of Permutation.vinigmat wrote:C will be to the right of D in exactly half of the cases. Therefore, you need to divide 3360 by 2.
Can somebody please explain the quote.Wht does C be right of D exactly half the cases.I got the first part of 3360
New here Create free account