In how many was can the letters of the word ELATION be arranged?

1. Analyze the question

The question asks for the number of ways to arrange the letters of the word ELATION. This is a problem of permutations, specifically dealing with the arrangement of distinct objects.

2. Explain the concept

When you are asked to arrange a set of objects (or letters), you use the concept of factorials.

• If all the items in a set are distinct (none of them repeat), the number of ways to arrange $n$ items is given by $n!$ (n factorial).
• $n! = n \times (n-1) \times (n-2) \times \dots \times 1$.

First, let's count the letters in the word ELATION: E, L, A, T, I, O, N. There are 7 letters in total.

Next, check for repetitions:

• E: 1
• L: 1
• A: 1
• T: 1
• I: 1
• O: 1
• N: 1 Since all 7 letters are unique (distinct), we do not need to divide by any repeating factorials.

3. Show the solution

• Total number of letters ($n$) = 7
• Since all letters are distinct, the number of arrangements is $7!$
• $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
• $7! = 5040$

Looking at the provided options, the question asks for the answer in factorial notation rather than the calculated numerical value.

4. Identify the answer

Comparing our result ($7!$) with the given options: A. 6! B. 7! C. 5! D. 8!

The correct option is B.

5. Tip

Always check for repeating letters first! If the word had been "BALL" (4 letters, with two Ls), the formula would be $\frac{4!}{2!}$. Because "ELATION" has no repeating letters, it is a straightforward $n!$ calculation.