Okay, so I need to figure out the number of ways to arrange the letters of the word ELLIPSE. Let me start by understanding the problem. The word is ELLIPSE. First, I should write down the letters and check how many times each letter repeats. 

Let me list them out: E, L, L, I, P, S, E. Wait, that's seven letters. Let me count again: E, L, L, I, P, S, E. Yep, seven letters. Now, let's check each letter's frequency. 

E appears twice. L appears twice. The other letters—I, P, S—each appear once. So, in total, the word has 7 letters with duplicates for E and L. 

I remember that when arranging letters with duplicates, the formula is the factorial of the total number of letters divided by the product of the factorials of the counts of each repeated letter. So, the formula should be 7! divided by (2! * 2!). Let me confirm that. 

Yes, because there are two E's and two L's. So, 7! is 5040, and 2! is 2, so 2! times 2! is 4. Then 5040 divided by 4 is 1260. So, the number of distinct arrangements should be 1260. 

Wait, but let me make sure I didn't miss any other duplicates. Let me check the word again: E, L, L, I, P, S, E. No, only E and L are repeated twice. The others are unique. So, the calculation should be correct. 

Alternatively, maybe I can think of it step by step. Suppose all letters were unique. Then there would be 7! ways. But since E is repeated twice, we have overcounted by 2!, and similarly for L. So, dividing by 2! for E and 2! for L gives the correct count. 

Another way to think about it: first choose positions for the E's, then the L's, and the rest will be unique. 

Number of ways to choose 2 positions out of 7 for E's: C(7,2). Then, from the remaining 5 positions, choose 2 for L's: C(5,2). The remaining 3 positions will be filled by the unique letters I, P, S in 3! ways. 

Calculating that: C(7,2) is 21, C(5,2) is 10, and 3! is 6. Multiplying them together: 21 * 10 * 6 = 1260. 

Same result. So that confirms the answer. 

Alternatively, using permutations with repetition: the formula is n! / (n1! * n2! * ... * nk!), where n is total letters, and n1, n2,...nk are the counts of each repeated letter. In this case, n=7, n1=2 (for E), n2=2 (for L), and the rest are 1. So, 7! / (2! * 2!) = 5040 / 4 = 1260. 

Therefore, the number of distinct arrangements is 1260. 

I think that's thorough. I considered different methods and they all lead to the same answer. So, confident that 1260 is the correct number of arrangements.
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