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In how many ways can we arrange the 13 letters of the word “COMMUNICATION” in which

 

(i}      there are no restriction.

 

(ii)    the word start with M and end with I.

 

(iii)    the two letters C do not occur next to each other.

 Oct 13, 2017
 #1
avatar+98196 
+2

Here's the first one

 

C - O - M - M - U - N - I  - C - A -  T - I - O - N

 

The number of  "identifiable" arrangements is given by :

 

13!  /  [ 2! * 2!  * 2!  * 2! * 2!  ]  =

 

13!  [ 32]  =  194,594,400  "words"

 

 

cool cool cool

 Oct 13, 2017
edited by CPhill  Oct 13, 2017
edited by CPhill  Oct 13, 2017
 #2
avatar+98196 
+2

Here's the third one :

 

Let's count the ways where the two C's can appear together

 

They can appear together in  any one of twelve positions

 

And the number of identifiable arrangements of the other letters  is

 

11 ! / [ 2! * 2! * 2! * 2! ]

 

So.....the total number of ways where they can occur together  is

 

12 * 11!  / [ 16]  = 29,937,600

 

So.....the number of ways in which they don't appear together is

 

194,594,400 - 29,937,600  =  164,656,800

 

 

cool cool cool

 Oct 13, 2017
edited by CPhill  Oct 13, 2017

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