How many ways to arrange?

Calculator: 10 letters

(a) repeats twice

(c) repeats twice

(l) repeats twice 

So, \(\frac{10!}{2!*2!*2!}=453600\)

Yes, like as what tertre said:

Here's a similar problem, How many ways are there to arrange the word, "Waterfall"

First, count how many letters, and see how many letters in the word repeat, then divide.

If the word did not have any repeating letters, then this problem would be much simpler. 

For example, let's think of the number of possible arrangements for the not-so-arbitrary word "Melody."

• For the first letter, there are 6 total choices
• For the second letter, there are now 5 possible choices
• For the third letter, there are now 4 possible choices
• This goes on until the final letter where there is only one option. 

Therefore, \(6*5*4*3*2*1\) is the total number of combinations. In shorthand, this can be mathematically expressed with a factorial symbol: \(6!\). 

If there are repetitions in the word, then simply divide by the number of arrangements of the repeated letters. 

Therefore, in "calculator," there are three repeated letters. There are:

• 2 "c's"
• 2 "a's"
• 2 "l's"

How many ways are there to arrange the word, "Calculator"

Count the letters in the word "Calculator":

\(\begin{array}{|c|r|} \hline \text{letter} & \text{count} \\ \hline c & 2 \\ a & 2 \\ l & 2 \\ u & 1 \\ t & 1 \\ o & 1 \\ r & 1 \\ \hline \text{sum} & \color{red}10 \\ \hline \end{array} \)

How many ways are there to arrange the word, "Calculator":

\(\begin{array}{|rcll|} \hline && \dfrac{ {\color{red}10} !}{2!2!2!1!1!1!1!} & | \quad 1! = 1 \\\\ &=& \dfrac{ {\color{red}10} !}{2!2!2!} & | \quad 2!=2 \\\\ &=& \dfrac{ {\color{red}10} !}{2\cdot 2\cdot 2} \\\\ &=& \dfrac{ {\color{red}10} !}{8} & | \quad 10! = 2\cdot 3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10 \\\\ &=& \dfrac{ {2\cdot 3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10} }{8} \\\\ &=& 2\cdot 3\cdot4\cdot5\cdot6\cdot7\cdot9\cdot10 \\\\ &=& 453600 \\ \hline \end{array}\)