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This question already has an answer here:

How many distict arrangements of the word $\text{MATHEMATICS}$ are there that contain no $A$'s in the first 7 spaces? I'm not quite sure how I would go about answering this. At first I thought I would calculate the number of arrangements in which $A$'s are in the first 7 spaces, and subtract it from the total number of distinct arrangements, but I have no idea what to do.

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marked as duplicate by José Carlos Santos, YuiTo Cheng, Shailesh, Cesareo, Jendrik Stelzner May 28 at 11:10

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In $\text{MATHEMATICS}$, we have total of $11$ letters, of which 2 are $A$s, 2 are $M$s, 2 are $T$s, and the rest are different.

Let's consider the last four spaces, as ArsenBerk and I said before.

So, We have $4$ spaces, and we need to put 2 $A$s in them.

So, Number of ways to do it : $^4C_2 = 6.$

[The reason I used combination as order of both $A$s does not matter.]

We still have $9$ more letters to care about, though.

We already have sorted the $A$s, and We have $9$ spaces remaining to fill up with $9$ letters, some of which are repeated.

Number of permutations = $$\frac{9!}{2!2!} = 90720$$

So, The total number of distinct arragements with no $A$s in first $7$ spaces = $6 * 90720 = 544320$

I guess I'm correct.

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HINT: Since there are no $A$'s in the first seven spaces, both of $A$'s should be in the last four spaces. If we choose the places of $A$'s first and then permute the other nine letters, what will be the answer?

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