$\begin{array}{1 1} 136 \\ 192 \\ 1680 \\ 2454 \end{array}$

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In the word $MATHEMATICS$ there are

$M$......2

$A$.......2

$T$.......2

H,E,I,C,S....5

Case (i) Four letters selected are :

Any two pairs of alphabets from M,A,T

The selection can be done in $^3C_2$ ways and then arranged in $\large\frac{4!}{2!.2!}$ ways.

Case (ii) Four letters selected are :

Two letters from any of the pairs of M,A,T and two from any of H,E,I,C,S and any 2 alphabets from M A,T other than the pair already selected.

The selection can be done in $^3C_1.^7C_2$ ways and

arranged in $\large\frac{4!}{2!}$

Case (iii) All the 4 letters are different.

The selection is done in $^8C_4$ ways and arranged in $4!$ ways.

$\therefore$ The required no. of arrangements

$=^3C_2.\large\frac{4!}{2!.2!}$$+^3C_1.^7C_2.\large\frac{4!}{2!}$$+^8C_4.4!$

$=3\times 6+3\times 21\times 12+70\times 24$

$=18+756+1680=2454$

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