Hard Question(Permutation)

[ChessWriter]

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

[shankar.ashwin]

8 letters with 2A's, and 3B's can be arranged in 8!/3!2! ways = 3360.

In half of the arrangements C would be to the right of D (or) 3360/2 = 1680

[pemdas]

8P8/(3!*2!)=3360 way the letter can be distributed
3 B and 2 A repeat

two positions around D, when the letter C is in front of and behind of the letter D.

3360/2=1680

a

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

[ChessWriter]

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.

[ronnie1985]

Since, the location of C wrt D is not specified, only it will be placed to right of D. The total no of ways the letters can be arranged is = 8!/(3!2!) = 3360 in which there will be 3360/2 ways in which C will be in the right of D.

[ArunangsuSahu]

As this is a permutation
C left of D +C Right of D=3360

So C Right of D=1680