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: \[179.\]
How many possible words can be created from the letters R, A, N, D (with repetition)?
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: