In how many ways can the word INITIAL be arranged?

I know that if all the letters were different (aka ABCDEFG) then the answer would be 7!, and I think if two of the letters were the same (aka AABCDEF) then the answer for be 7!/2! (or maybe 7!/2?), but I'm not sure what the answer would be be for three letters being the same (aka AAABCDE or IIINTAL)

For

ABCDE, it's 5!
AAACDE, it's 6!/3!
AABBCC, it's 6!/(2!2!2!)

To calculate the number of ways the word INITIAL can be arranged, we need to consider the repetitions of certain letters.

1. When all the letters are different (e.g., ABCDEFG), the number of arrangements is given by the factorial of the number of letters, which is 7! = 5040.

2. When two letters are repeated (e.g., AABCDEF or IIINTAL), we need to divide the total number of arrangements by the factorial of the number of repetitions. In this case, since there are two A's in the word INITIAL, we divide by 2!. So, the number of arrangements would be 7! / 2! = 7 * 6 * 5 * 4 * 3 = 840.

3. When three letters are repeated (e.g., AAABCDE), we need to divide the total number of arrangements by the factorial of the number of repetitions. In this case, since there are three A's in the word INITIAL, we divide by 3!. So, the number of arrangements would be 7! / 3! = 7 * 6 * 5 * 4 = 840.

Therefore, the number of ways the word INITIAL can be arranged in total is 840 when there are two or three repetitions of a letter.

To find the number of ways the word INITIAL can be arranged, we need to consider the repetition of letters.

Let's break down the word INITIAL and analyze the repeating letters:
- There are two 'I's in the word INITIAL.
- All other letters are unique.

If all the letters were different, as in ABCDEFG, then you are correct in saying that there would be 7! (7 factorial) ways to arrange them.

In the case of two identical letters, like AABCDEF, we need to account for the repetition. The formula for finding the number of arrangements with repeated elements is to divide the factorial of the total number of elements by the factorial of the repeated elements. So, there would be 7!/2! (or 7!/2) ways to arrange the letters AABCDEF.

For three identical letters, like AAABCDE or IIINTAL, we apply the same concept. We divide the factorial of the total number of elements by the factorial of the repeated elements. Therefore, there would be 7! / (3!) ways to arrange the letters with three 'A's or three 'I's.

To calculate the number of arrangements for the word INITIAL, we need to consider that there are two 'I's in total. Therefore, we divide the total number of arrangements (7!) by the factorial of the repeated elements (2!). The answer is:

7! / 2! = 7 * 6 * 5 * 4 * 3 * 2 * 1 / (2 * 1) = 5040 / 2 = 2520

So, the word INITIAL can be arranged in 2520 different ways.