Probability - Permutation Q

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 would be 8! / 2! x 3!

Can someone please explain in detailed steps? Thanks in advance!

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

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.

[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]

What is the source of this question?

To me, the answers given only work if C is adjacent to D. This isn't specified in the question.

*edit* actually I just read the second half of Sanju's post - ignore my comment

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!

It will be better to break the problem by omitting the repeating letters.
How many ways can ABCD be arranged so that B is always immediately to the right of C.
Consider B and C as one letter say X... so you are left with three letters A X D
Arranging the letters is possible in 3! ways.
But the letter B & C are actually two letters , they can be arranged among themselves in 2 ways
BC or CB , u want only the first arrangement so only one arrangement is possible.
Notice I have said immediately to make it simpler.

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

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

This is a very important concept of Permutation.

Take for example P Q R

WE can arrange P Q R in !3 ways. = 6 ways

P Q R

P R Q

R P Q

R Q P

Q P R

Q R P

You can see At the First Place exactly 2 times each Letter comes , P - 2 , Q - 2 , R - 2

Now if we ask to calculate --- In total 6 ways,-------- Any letter can come from 3 ,,,, so 6 / 3 = exactly 2 times....

How many times Q Will be exactly right to P ----

Ist CASE
Q will be right to P....

IInd CASE
Q will not be right to P...

So 6 / 2 = 3
In total 3 times Q will be right to P. ...

To be precise ...

Total number of out comes / Total number of cases.

In First example 6/3 Because either P, Q or R can come ---- so in total 3 cases ...

In second example 6/2, Now its only 2 cases. 6/2 = 3

Its not easy to understand at first, Because question is constructed in a way, It looks a lot confusing....Keep Trying & Hope for the Best...........