How many ways can we rearrange the letters in DISCRETE

  • Thread starter ZombiesFTW
  • Start date
  • #1

Main Question or Discussion Point

Hey, new member here. Been viewing this forums for a long time now and used this forum as a resource for help with homework.

Anyways, wasn't sure where to put this problem. I am having trouble setting up the problem.

How many ways can we rearrange the letters in DISCRETE so that the E’s are adjacent and the
first letter precedes the last letter in the standard alphabetical order (e.g. DISCREET is ok, but
TISCREED is not).

I am trying to solve it by realizing there are 8 spaces in that word. Then how many different places can I place the E's so then that would be something like this:

_ _ _ _ _ _ _ _
8*8*6*5*4*3*2*1
6*8*8*5*4*3*2*1
.
.
.
.
.
+_________________
Add up each row's product and that would be the answer.

Is this the best way to go about solving this? Or is there an easier way?
 

Answers and Replies

  • #2
34
0
I think the easiest way to do it is consider the two E's as a single letter thus your words are built from D I S C R EE T - thus you have 7 distinct "letters".

As for have the ordering between the first and last letter, you can ignore this and calculate all possible words then divide by 2 to only consider the correct order. The reason for this factor of two is that all words come in pairs: A word and its reverse. Only one of these will have the ordering between the first and last letter you want.
 
  • #3
So then would the answer be 7! / 2?

or

Would it be (7! * 7) / 2

Since it would be
7*6*5*4*3*2*1
6*7*5*4*3*2*1
6*5*7*4*3*2*1
.
.
.
.
6*5*4*3*2*1*7
+_______________
7! * 7
then divide by two since the order would only work one way.
 
Last edited:
  • #4
590
0
7!/2 is correct, I don't quite understand how you got 7*7!
 

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