SOLUTION: How many arrangements of the word ACTIVE are there if the C and E a) must always be together? b)must never be together?

(a)  If the C and E are always together, we can consider this pair as one merged object.

     Then we have 5 objects in total that admit 5! = 1*2*3*4*5 = 120 permutations.

     But the merged object has two states: CE and EC.

     By combining all these permutations, we have, in total, 2*5! = 2*120 = 240 permutations.

     It is the ANSWER in this case.



(b)  This time, we should consider all permutations of 6 letters of the word ACTIVE, 6! = 720 permutations, 

      and subtract those 2*120 = 240 permutations, found in the case (a), where the letters C and E are placed together.


      So, the ANSWER  in this case is  6! - 2*5! = 720 - 240 = 480.