In how many different ways can you order in line the letters of the word MONADOLOGY? In how many different ways can you order in line the letters of the word MONADOLOGY?
I was thinking - because of the 3 O's, each permutation repeats $3!$ times. Because of that the answer should be $10!$ (all the ways to order the letters) divide by $3!$
What do you think?
 A: Reworded, you are asking how many strings can be constructed using letters from the multiset $\{A,D,G,L,M,N,O,O,O,Y\}$ where each letter is used exactly as many times as it appears in the multiset.
For notation, we often write each element only once, but will write a number to the left of each representing the number of times it appears in the multiset.  For your case, it is $\{1\cdot A, 1\cdot D, 1\cdot G, 1\cdot L, 1\cdot M, 1\cdot N, 3\cdot O, 1\cdot Y\}$
In general, with multiset $\mathcal{M} = \{\alpha_1\cdot A_1, \alpha_2\cdot A_2,\dots, \alpha_n\cdot A_n\}$ where $|\mathcal{M}|=\alpha_1+\alpha_2+\dots+\alpha_n = N$,
the number of multipermutations will be $\frac{N!}{\alpha_1!\alpha_2!\dots\alpha_n!}$
This is such a commonly occurring problem that we give it its own notation:
$\binom{N}{\alpha_1,\alpha_2,\dots,\alpha_n}$
For your specific case, you have ten letters total, seven of which occurring once and one of which repeated a total of three times, so the total will be:
$\binom{10}{1,1,1,1,1,1,1,3}=\frac{10!}{1!1!1!1!1!1!1!3!}=\frac{10!}{3!}$

For additional example, the quintessential example is "How many arrangements of the letters of the word MISSISSIPPI exist?"
The multiset being $\{4\cdot I, 1\cdot M, 2\cdot P, 4\cdot S\}$
The number of multipermutations being $\binom{11}{4,1,2,4} = \frac{11!}{4!1!2!4!}$

The proof of the formula follows from induction and direct application of multiplication principle following the steps:


*

*Pick which of the $N$ locations are occupied by the $\alpha_1$ copies of $A_1$

*Pick which of the remaining $N-\alpha_1$ locations are occupied by the $\alpha_2$ copies of $A_2$

*Pick which of the remaining $N-\alpha_1-\alpha_2$ locations are occupied by the $\alpha_3$ copies of $A_3$

*$\dots$

