Find the number of permutations of the letters in the word, MONOTONOUS. In how many ways are four O's together? In how many ways are (only) 3 O's together?

In combinatorial analysis, permutations with repetition allow us to determine the number of unique arrangements of a set where some elements are repeated. This scenario often occurs with words or sets containing duplicate items.
For example, in the word MONOTONOUS, the letter 'O' repeats four times. To calculate permutations correctly, we need to account for these repetitions. This is achieved using the formula:
\[ \frac{n!}{n_1! \times n_2! \times \text{...} \times n_k!} \] where \( n \) is the total number of items, and \( n_1, n_2, \text{...}, n_k \) are the frequencies of the repeated items.
So for MONOTONOUS:
\[ \frac{10!}{4! \times 2! \times 2!} \] Here, we divide 10! by the factorials of the counts of each unique letter ('O' repeated 4 times, 'N' repeated twice, 'O' repeated twice). This method ensures each identical letter arrangement isn't overcounted.

A factorial of a number (\texttt{n!}) is the product of all positive integers from 1 up to that number. It forms the foundation for many combinatorial calculations, including permutations and combinations.
Let's consider the factorial calculation for the word MONOTONOUS:
- First, calculate the total number of letters, which is 10. Thus, \( 10\texttt{!} = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800 \).
- Then, for the repeated letters 'O' (4 times), 'N' (2 times), and 'O' (2 times), calculate their factorials: \( 4\texttt{!} = 24, 2\texttt{!} = 2 \).
- Finally, plug these into our permutation formula:
\[ \frac{10!}{4! \times 2! \times 2!} = \frac{3,628,800}{24 \times 2 \times 2} = 37,800 \]
Calculating factorials accurately is crucial. Even a small mistake can lead to incorrect results, especially in larger sets.

Combinatorics often involves solving word problems to find various permutations and combinations. These exercises help understand the practical applications of the mathematical concepts.
Let's break down the given exercise about the word MONOTONOUS:
- When 4 'O's are together, treat them as a single entity. This changes the set from 10 letters to 7 entities (4 'O's + 6 unique letters).
- Now find the permutations with repetitions for these 7 entities:
\[ \frac{7!}{2! \times 2!} = \frac{5,040}{4} = 1,260 \]
- Similarly, when only 3 'O's are together, treat 3 'O's as one entity. This gives us 8 entities (3 'O's + 1 'O' + 6 unique letters):
\[ \frac{8!}{3! \times 2! \times 2!} = \frac{40,320}{48} = 840 \]
Combinatorial problems in words require breaking down sets into manageable entities and applying permutation formulas with precise calculations. Consistent practice with different problems helps build a deeper understanding and ensure accuracy.