# Number of distinct arrangements of the word $\text{MATHEMATICS}$ [duplicate]

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.

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

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.

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?