Awli wrote:

How many different possible arrangements can be obtained from the letters G, M, A, T, I, I, and T, such that there is at least one character between both I's?

A. 360

B. 720

C. 900

D. 1800

E. 5040

There are 7 letters G, M, A, T, I, I, and T with I and T repeated twice.

Hence total arrangements/permutations of letters = 7!/(2! * 2!) = 1260.

Now, let us say that the two "I" are always together.

So now we have 6 letters G, M, A, T, II, T with T repeated twice.

Hence total arrangements/permutations of letters = 6!/(2!) = 360.

Required arrangements = 1260 - 360

= 900

Hence option (C).

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