Question

In how many ways can the 7 letters A,B,C,D,E,F and G be arranged so that C and E never together.
(a) 5040
(b) 6480
(c) 3600
(d) 1440

Answer 1

Answer:

3600

Step-by-step explanation:

let  assume that CE is a single letter

Total possible arrangement : 7!

Arrangements where C & E are together: 2x6!

So, arrangements where C&E are not together:  

7! - 2x6!

= 7x6! -2x6!

= 5x6! = 5x 720= 3600

Answer 2

Answer:

The correct answer is option(c) 3600.

Step-by-step explanation:

Given:

The 7 letters A, B, C, D, E, F, and G be arranged. so that C and E are never together.

To find:

How many ways can the 7 letters A, B, C, D, E, F, and G be arranged so that C and E are never together?

First, organize these seven letters without any restrictions. So the entire number of methods to do this is $7!$ as we have entire seven letters.

Now, according to given circumstances we have to create words that don't have C&E together. To solve this assume that C&E is one unit, now the total number of words becomes 6! as we have left with 6 letters.

Now C&E are convertible so now again we include 6! ways to organize them.

So the needed solution is

$=7!-6!-6!$

$=7*6*5*4*3*2*1-2(6*5*4*3*2*1)$

$=5040-2(720)$

$=5040-1440$

$=3600$

So there exist $3600$ words that can be made according to given circumstances.

Therefore, the correct answer is option (c) 3600.