[email protected] wrote: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
Can you please explain the above in detail. Appreciate.
Epitome: Do not permute (DC).
Meticulous account: The arrangements will have the letter C to the right of the letter D, like
+ + + + D C + + etc
In other words, C and D are occurring together in an unmoved manner. We'll take them together as a single item and their mutual arrangements won't be considered.
Also, know that the number of permutations of n things out of which p are alike and are of one type, q are alike and are of a second type, r are alike and are of a third type, and the rest are all different
= n! / (p! q! r!)
The present case has only two types that repeat that are 2 A_s and 3 B_s, and the rest C, D, and E are all different. When (DC) is taken as single as it is, we're left with just 7 letters that include 2 A_s and 3 B_s as well. Total permutations in such case
= 7! / (2! 3!)
We would have needed to further multiply that by 2! in order to consider DC occurring both ways (DC & CD) in all permutations, but only one of the two ways, i.e. DC, is considered, and the correct answer may well be 7! / (2! 3!) = 420.
The language of the question is making me believe that the letter C must be to the immediate right of the letter D, and so is my answer. But after reading through the choices, I learnt that the letter C must be anywhere to the right of the letter D. For each arrangement of the type + D + + + C + +, there always is an arrangement of the type + C + + + D + +, which we're not supposed to include. So we can safely take half of the total permutations in that case, which is =
½ × 8! / (2! 3!) = [spoiler]
1680.
A[/spoiler]
