Permutations of word 'mathematics' How many arrangements are there of MATHEMATICS with both T's before both A's or both A's before both M's or both M's before the E ?
Can someone also point to some online resource that has such practice questions ?
Thanks.
Edit:
I have some initial thought, please tell me if I am in the right direction :
For the case both T's before both A's :
We can picture the permutation as follows :
(7) T (7) T (7) A (7) A (7)  (Remaining letters: M H E M I C S - total 7)
Each of the brackets are empty spaces for now. We need to pick 7 out of the total 7 X 5 = 35 spaces; that will ensure both T's being before both A's.
So number of ways = C(35,7) X 7!/2! (because two M's are same - we need to divide by 2!)
 A: You were headed in the right direction, but you went way too far.  The $C(35,7)$ is much too much.  It counts, for example, "-M---I-T..." and "--M-I--T..." as different, even though both are basically "MIT..."
Here's a way to think of it.  Starting from TTAA, allow yourself to insert the other letters, C, E, H, I, M, M, and S, one at a time, anywhere before, between, or after letters already in position.  (It might help to picture the second M as, temporarily, an N, remembering to divide by $2$ when you're done.)  The C have $5$ places it can go, the E then has $6$, the H has $7$, and so forth, for a total of
$$(5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11)/2$$
Can you now do the other examples you asked about (both A's before both M's or both M's before the E)?  The first of those should actually be very easy!
A: First the seven letters _M_H_E_M_I_C_S_ can be arranged among themselves in $7!/2!$ ways.
Then we are left with $8$ places in between, infront or at the end of word.
Now we have to Insert two A's at the last $6$ gaps this can be done in $C(6,2)/2!$ ways.  After filling in these gaps we are left with $6$ gaps which can be filled by remaining two T's in $C(6,2)/2!$ ways.
Hence the required number of ways  is $$\frac{7! \cdot C(6,2) \cdot C(6,2)}{2! \cdot 2! \cdot 2!}$$ 
Hope you understand this.
A: A particular permutation of the letters of the word MATHEMATICS is determined by choosing which two of the eleven places will be filled with M's, then choosing which two of the remaining nine places will be filled with A's, then choosing which two of the remaining seven places will be filled with T's, and finally permuting the five remaining letters.  Therefore, the number of permutations of the letters of the word MATHEMATICS is 
$$\binom{11}{2}\binom{9}{2}\binom{7}{2} \cdot 5! = \frac{11!}{9!2!} \cdot \frac{9!}{7!2!} \cdot \frac{7!}{5!2!} \cdot 5! = \frac{11!}{2!2!2!}$$ 
There are 
$$\binom{4}{2} = 6$$
permutations of the letters AATT since a particular permutation is determined by selecting which two of the four places will be filled with A's.  Of these six permutations, only one (TTAA) has both T's before both A's.  Thus, the number of permutations of MATHEMATICS in which both T's appear before both A's is 
$$\frac{1}{6} \cdot \frac{11!}{2!2!2!}$$   
A: I tried : Let A,B,C are sets  respectively  in which arrangement of MATHEMATICS in which both T's before both both A's ,both A's before both M's ,both M's before the E.  Now we solve this problem by using set theorotical formula  :-   |A union B union C|= {|A|+|B|+|C|} - {|A intersection B| +|B intersection C|+|A intersection C|} + { |A intersection B  intersection C |}  So the answer is : {C(11,4)×7!/2! + C(11,4)×7!/2! + C(11,3)×8!/2!2!} - { C(11,6)×5!+ C(11,5)× 6!/2!+ C(11,4)×C(7,3)×4! } + { C(11,7)×4! }
