+0

# probability

0
92
2

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.

Guest Oct 13, 2017
Sort:

#1
+79881
+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"

CPhill  Oct 13, 2017
edited by CPhill  Oct 13, 2017
edited by CPhill  Oct 13, 2017
#2
+79881
+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

CPhill  Oct 13, 2017
edited by CPhill  Oct 13, 2017

### 6 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details