To find the number of distinct arrangements of the letters in "INITIATION", we can use the formula for permutations of multiset: n! / (r1! * r2! * ... * rk!), where n is the total number of items, and r1, r2, ..., rk are the numbers of each type of item.

In "INITIATION", there are 10 letters in total.

The letter 'I' appears 3 times, 'N' appears 2 times, 'T' appears 2 times, and the letters 'A' and 'O' each appear once.

So, the number of distinct arrangements is:

10! / (3! * 2! * 2! * 1! * 1!) = 3,628,800 / (6 * 2 * 2 * 1 * 1) = 151,200.

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To find the number of distinct arrangements of the letters in "INITIATION", we can use the formula for permutations of multiset: n! / (r1! * r2! * ... * rk!), where n is the total number of items, and r1, r2, ..., rk are the numbers of each type of item.

In "INITIATION", there are 10 letters in total.

The letter 'I' appears 3 times, 'N' appears 2 times, 'T' appears 2 times, and the letters 'A' and 'O' each appear once.

So, the number of distinct arrangements is:

10! / (3! * 2! * 2! * 1! * 1!) = 3,628,800 / (6 * 2 * 2 * 1 * 1) = 151,200.

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To find the number of distinct arrangements of the letters in "INITIATION", we can use the formula for permutations of multiset: n! / (r1! * r2! * ... * rk!), where n is the total number of items, and r1, r2, ..., rk are the numbers of each type of item.

In "INITIATION", there are 10 letters in total.

The letter 'I' appears 3 times, 'N' appears 2 times, 'T' appears 2 times, and the letters 'A' and 'O' each appear once.

So, the number of distinct arrangements is:

10! / (3! * 2! * 2! * 1! * 1!) = 3,628,800 / (6 * 2 * 2 * 1 * 1) = 151,200.

Find the number of distinct arrangements of the 10 letters in INITIATION.Two of the same letter are considered identical (not distinct).

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