In how many distinguishable ways can the seven letters in the word MINIMUM be arranged, if all the letters are used each time?

Options

A. 7
B. 42
C. 420
D. 840

Explanation

To find the number of distinguishable ways to arrange the letters in the word "MINIMUM," we need to consider the repetitions of the letters. Here's the step-by-step solution:

### Step 1: Identify the letters and their frequencies
The word "MINIMUM" consists of 7 letters. Let's break down the letters:
- M appears 3 times
- I appears 2 times
- N appears 1 time
- U appears 1 time

### Step 2: Use the formula for permutations of multiset
The formula for the number of distinguishable arrangements of letters when there are repetitions is given by:

\[ \text{Number of arrangements} = \frac{n!}{n_1! \times n_2! \times n_3! \times \ldots} \]

where:
- \( n \) is the total number of letters,
- \( n_1, n_2, n_3, \ldots \) are the frequencies of the repeated letters.

In our case:
- Total letters \( n = 7 \)
- Frequencies: M = 3, I = 2, N = 1, U = 1

### Step 3: Substitute the values into the formula
Now we substitute the values into the formula:

\[ \text{Number of arrangements} = \frac{7!}{3! \times 2! \times 1! \times 1!} \]

### Step 4: Calculate the factorials
Now we calculate the factorials:
- \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \)
- \( 3! = 3 \times 2 \times 1 = 6 \)
- \( 2! = 2 \times 1 = 2 \)
- \( 1! = 1 \)

### Step 5: Substitute the calculated factorials back into the equation
Now we substitute these values back into our equation:

\[ \text{Number of arrangements} = \frac{5040}{6 \times 2 \times 1 \times 1} = \frac{5040}{12} \]

### Step 6: Perform the division
Now we perform the division:

\[ \frac{5040}{12} = 420 \]

### Final Answer
Thus, the number of distinguishable ways to arrange the letters in the word "MINIMUM" is **420**.

---