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# permutations

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601
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how many permutations in this word?

COMBINATORICS

math statistics
Aug 21, 2014

#2
+100256
+10

Anonymous appears to be correct except that there are 13 letters.

(Unless I can't count which is always a possibility)

$$Number of Permutations is \frac{13!}{2!2!2!}$$$$${\frac{{\mathtt{13}}{!}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}\right)}} = {\mathtt{778\,377\,600}}$$ I am never overly confident about these so I used this site for reference. http://www.regentsprep.org/regents/math/algebra/apr2/LpermRep.htm Aug 22, 2014 ### 2+0 Answers #1 +5 So we have the word "combinatorics" which can be rearranged to get abcciimnoorts. It is a 12 letter word so we start with 12! which is 479001600 although this assumes all the double letters are unique. We then need to divide by the total number of permutations of the double letters c, I, and o. The total number for this is 2!2!2! which is 8, and so we get 479001600/8 which becomes 59875200. Aug 21, 2014 #2 +100256 +10 Best Answer Anonymous appears to be correct except that there are 13 letters. (Unless I can't count which is always a possibility) $$Number of Permutations is \frac{13!}{2!2!2!}$$$

$${\frac{{\mathtt{13}}{!}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}\right)}} = {\mathtt{778\,377\,600}}$$

I am never overly confident about these so I used this site for reference.

http://www.regentsprep.org/regents/math/algebra/apr2/LpermRep.htm

Melody Aug 22, 2014