Question #36053

1 Answer
Jan 5, 2018

#(12!)/2^4=29936700#

Explanation:

It might be tempting to just say #12!#, since that is the length of the word, but this is not the case. If all the letters were only in the word once, that would be the case, but we have several duplicates.

To figure out a method to dealing with this, let's take a simpler example. Suppose we wanted to figure out in how many ways the word #EGG# could be arranged. If we just straight up took the factorial, we would count all these as arrangements:
#EGG#
#EGG#
#GEG#
#GEG#
#GGE#
#GGE#

We can see that we have counted all of the permutations twice. This is because the factorial count treats #EGG# with the first #G# coming first different from #EGG# with the second #G# coming first, but to us, they're clearly the same.

We can solve this problem by dividing by #2!# to get rid of all the the double-counted arrangements.

In fact, you can generalize to say that if you have a letter occur #n# times in the word, then you have to divide the original factorial arrangements by #n!#, to remove all the duplicates.

In the case of #AR RANGEMENTS#, there are 2 of #A'#s, #R'#s, #N'#s and #E's#, which means we should divide by #2!# four times. This makes the total number of possibilities:
#(12!)/(2!*2!*2!*2!)=(12!)/2^4=479001600/16=29936700#