1. Count the letters available in MATHEMATICS

   The word MATHEMATICS has the following letter multiplicities:

   • $M$ appears $2$ times
   • $A$ appears $2$ times
   • $T$ appears $2$ times
   • and $H,E,I,C,S$ each appear $1$ time

   So there are $8$ distinct types of letters: $M,A,T,H,E,I,C,S$

2. Interpret the question

   We must choose $5$ alphabets from these letters, with the condition that chosen alphabets are not necessarily distinct, but we cannot exceed the available multiplicity in the word.

   So we need the number of nonnegative integer solutions of $m+a+t+h+e+i+c+s=5$ subject to $m,a,t\le 2, \qquad h,e,i,c,s\le 1.$

3. Casewise counting based on repetitions

   Since only $M,A,T$ can repeat, possible patterns for choosing $5$ letters are:

   Case 1: All $5$ chosen letters are distinct

   We choose any $5$ distinct letter-types from the $8$ available: $\binom{8}{5}=56$

   Case 2: Exactly one letter is repeated twice

   The repeated letter must be one of $M,A,T$.

   • Choose the repeated letter: $\binom{3}{1}=3$
   • Choose the remaining $3$ distinct letters from the other $7$ letter-types: $\binom{7}{3}=35$

   Total: $3\cdot 35=105$

   Case 3: Exactly two letters are repeated twice each

   Then the pattern is $2+2+1$.

   The two repeated letters must be chosen from $M,A,T$: $\binom{3}{2}=3$

   Then choose the remaining single letter from the other $6$ letter-types: $\binom{6}{1}=6$

   Total: $3\cdot 6=18$

4. Add all cases

   $56+105+18=179$

5. Conclusion

   The required number of ways is $\boxed{179}$

6. Option check

   • A: $179$ ✅
   • B: $177$ ❌
   • C: $175$ ❌
   • D: $181$ ❌

Therefore, the correct option is A.