Permutation of the swedish word "matematik" I have three questions and (I think) I have solved two of them so far. First one is

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*Calculate the number of words from the word "matematik" with switching order of the letters

this I got to be 45360 as its $\frac{9!}{2!2!2!}$


*calculate how many of these words contain both a $t$ in the end AND the start

this I got to $\frac{7!}{2!2!}$ or $1260$


*How many of these words contain the sequence "mat" atleast one place in the word

I do not really know how to approach this as it is 2 sequences of "mat" that can be made from the word "matematik" since m, a, t appear twice.
 A: The number of words that contain at least one "mat" may be evaluated by treating one set of the three constituent letters as a "macro letter". Then all $7$ letters are different and the number of admissible words is $7!$.
From this, the inclusion/exclusion principle says we must subtract the number of words with two "mat"s. Here, both macros "mat" are the same, so the number of words is $\frac{5!}2$. Subtracting gives the result as $7!-\frac{5!}2=4980$.
A: Q3
Easiest to over-count and substract back.
"mat" can start in any one of 7 places.  Once it starts, the other 6 letters can be in any order.
$A_1 = 7 \times 6! = 7!$
"mat...mat" possibilities re above have each been counted twice, so must be enumerated and deducted.
if 1st "mat" starts on position 1, 4 possibilites for 2nd "mat" to start.
if 1st "mat" starts on position 2, 3 possibilites for 2nd "mat" to start.
if 1st "mat" starts on position 3, 2 possibilites for 2nd "mat" to start.
if 1st "mat" starts on position 4, 1 possibilites for 2nd "mat" to start.
Total of 10 "mat...mat" placements.  With each placement
need 3! factor re (again) permuting 3 odd letters.
$A_2 = 10 \times 3!.$
Final answer = $A_1 - A_2.$

Addendum
Per OP's request.
Explanation for various methods of counting the # of ways that the "mat...mat..."
string can occur.
Once you compute the # of ways that a "...mat..." string can occur, you then have to
deduct the # of ways that the "mat...mat..." string can occur.  This deduction
is needed because in the original computation, each of these
"mat...mat..." occurrences was double-counted.
The (kludgy) approach that I took was to reason that the 1st "mat" string
would have had to begin somewhere in positions 1 through 4.  I then manually
determined that there were 10 different possible placements of "mat...mat...".
I then reasoned that each placement must be multiplied by $3!$, because the
three odd letters can permute.
Parcly Taxel also concluded that these "mat...mat..." placements must be
enumerated to compensate for their being overcounted.
His (elegant) approach to counting them was totally different than mine.
He reasoned as follows:
Suppose, instead of considering that you have 9 letters, you pretend that
you have 5 units, as listed below:
Unit-1 = "mat" 
Unit-2 = "mat" 
Unit-3 = "e" 
Unit-4 = "i" 
Unit-5 = "k"
Then you should also pretend that instead of having 9 letter-positions
you have 5 - unit-positions.
Then, you surmise that these 5 "units" can be permuted in any one of $5!$ ways
among these 5 unit-positions.
Then, you recognize that because Unit-1 and Unit-2 are identical, you have
(ironically) double-counted the # of possible distinct placements of these
5 units.
For example, the placement
Unit-1, Unit-2, Unit-3, Unit-4, Unit-5
has been counted separately from
Unit-2, Unit-1, Unit-3, Unit-4, Unit-5
This means that the correct # of distinct ways of permuting these 5 units,
given that Unit-1 and Unit-2 are identical is
$$\frac{5!}{2}.$$
So, you have two completely different (but both valid) approaches to counting
how many "mat...mat..." placements must be deducted.
My way was $10 \times 3!.$  His way was $\frac{5!}{2}.$
