How many 5 letter words can be formed from the letters of the word INDEPENDENCE

To determine how many 5-letter words can be formed from the letters of the word "INDEPENDENCE," we first need to analyze the letters and their frequencies. The word "INDEPENDENCE" consists of the following letters:

• I: 1
• N: 3
• D: 2
• E: 2
• P: 1
• C: 1

In total, there are 12 letters in "INDEPENDENCE," but since some letters repeat, we need to consider different cases based on the number of times each letter appears. Let's break it down into cases based on the number of unique letters used in the 5-letter combinations.

Case 1: All letters are different

In this case, we can choose 5 different letters from the available unique letters: I, N, D, E, P, C. There are 6 unique letters, and we can select 5 from these. The number of ways to choose 5 letters from 6 is given by the combination formula:

C(n, r) = n! / (r!(n - r)!)

Here, n = 6 and r = 5:

C(6, 5) = 6! / (5! * 1!) = 6

Now, for each selection of 5 letters, we can arrange them in 5! (5 factorial) ways:

5! = 120

Thus, the total for this case is:

6 * 120 = 720

Case 2: One letter appears twice, and three letters are different

In this scenario, we can have either N, D, or E appearing twice, as these are the only letters that can repeat. Let's analyze each sub-case:

• **N appears twice**: Choose 3 from {I, D, E, P, C} (5 options).
• **D appears twice**: Choose 3 from {I, N, E, P, C} (5 options).
• **E appears twice**: Choose 3 from {I, N, D, P, C} (5 options).

For each of these sub-cases, the number of ways to choose 3 letters from 5 is:

C(5, 3) = 10

Now, for each selection, we can arrange the letters (2 of one kind and 3 of different kinds) using the formula for permutations of multiset:

5! / (2!) = 120 / 2 = 60

Thus, for each of the three letters (N, D, E), the total arrangements are:

10 * 60 = 600

Since there are three letters that can appear twice, the total for this case is:

3 * 600 = 1800

Case 3: Two letters appear twice, and one letter is different

In this case, we can only have N and D, N and E, or D and E appearing twice. Let's analyze:

• **N and D appear twice**: Choose 1 from {I, E, P, C} (4 options).
• **N and E appear twice**: Choose 1 from {I, D, P, C} (4 options).
• **D and E appear twice**: Choose 1 from {I, N, P, C} (4 options).

For each of these combinations, the number of ways to arrange the letters (2 of one kind, 2 of another, and 1 of a different kind) is:

5! / (2! * 2!) = 120 / 4 = 30

Thus, for each of the three combinations, the total arrangements are:

4 * 30 = 120

So, the total for this case is:

3 * 120 = 360

Final Calculation

Now, we can sum up all the cases to find the total number of 5-letter words that can be formed:

• Case 1: 720
• Case 2: 1800
• Case 3: 360

Total = 720 + 1800 + 360 = 2880

Therefore, the total number of 5-letter words that can be formed from the letters of "INDEPENDENCE" is 2880.