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?

Question 1159623: How many arrangements of the word ACTIVE are there if the C and E
a) must always be together?
b)must never be together?
Answer by ikleyn(52780) (Show Source): You can put this solution on YOUR website!
.


(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.

Solved, answered, explained and completed.

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For many other similar solved problems, see the lesson
- Problems on Permutations with restrictions
in this site.

Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic "Combinatorics: Combinations and permutations".

Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.

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