Question

The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to :

• A. 175
• B. 181
• C. 177
• D. 179

Answer 1

The Correct Option is D

AA, MM, TT, H, I, C, S, E

(1) All distinct

8C5 → 56

(2) 2 same, 3 different

3C1 × 7C3 → 105

(3) 2 same 1^st kind, 2 same 2^nd kind, 1 different

3C2 × 6C1 → 18

Total → 179

Answer 2

Consider the distinct letters in the word MATHEMATICS: \( M (2), A (2), T (2), H (1), E (1), I (1), C (1), S (1) \). We aim to select 5 letters under different conditions of repetition.

Case 1: All five chosen letters are distinct. We choose 5 distinct letters from 8 available distinct letters:

\[\binom{8}{5} = 56 \text{ ways.}\]

Case 2: Two letters are the same, and three other letters are distinct. We first choose 1 letter to repeat from the letters M, A, or T (3 choices). Then, we choose 3 more distinct letters from the remaining 7: 

\[\binom{3}{1} \times \binom{7}{3} = 3 \times 35 = 105 \text{ ways.}\]

Case 3: Two letters of one kind are repeated, and two letters of another kind are repeated, with one additional distinct letter. We first select 2 letters to repeat from M, A, or T (choose 2 out of 3). Then, we select 1 distinct letter from the remaining 6: 

\[\binom{3}{2} \times \binom{6}{1} = 3 \times 6 = 18 \text{ ways.}\]

Summing all the cases gives: \[56 + 105 + 18 = 179 \text{ ways.}\]

Therefore: \[179.\]