The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is $ 4 \_\_\_\_\_$.
We are tasked with finding the total number of words that can be formed using 6 letters, where the letters are chosen from a set of 5 distinct letters. The solution is divided into three cases:
(i) Single letter is used:
If only one letter is used, then all 6 positions in the word must consist of that single letter. Since there are 5 distinct letters to choose from, the number of such words is:
$ 5 $
(ii) Two distinct letters are used:
If two distinct letters are used, we need to calculate the number of ways to form words under this condition. First, choose 2 letters from the 5 available letters:
$ {^5C_2} = 10 $
Next, distribute the 6 positions among the two chosen letters. There are two subcases:
- One letter appears twice, and the other appears four times.
- Both letters appear three times each.
For the first subcase (one letter appears twice, the other appears four times):
The number of arrangements is:
$ \frac{6!}{2!4!} = 15 $. Since either letter can appear twice, multiply by 2:
$ 15 \times 2 = 30 $
For the second subcase (both letters appear three times each):
The number of arrangements is:
$ \frac{6!}{3!3!} = 20 $
Thus, the total number of words for this case is:
$ 10 \times (30 + 20) = 10 \times 50 = 500 $
(iii) Three distinct letters are used:
If three distinct letters are used, we need to calculate the number of ways to form words under this condition. First, choose 3 letters from the 5 available letters:
$ {^5C_3} = 10 $
Next, distribute the 6 positions equally among the three chosen letters, with each letter appearing twice. The number of arrangements is:
$ \frac{6!}{2!2!2!} = 90 $
Thus, the total number of words for this case is:
$ 10 \times 90 = 900 $
Total Number of Words:
Adding up all the cases, the total number of words is:
$ 5 + 500 + 900 = 1405 $
Final Answer:
The total number of words is $ \boxed{1405} $.
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.
Let \( f : (0, \infty) \to \mathbb{R} \) be a twice differentiable function. If for some \( a \neq 0 \), } \[ \int_0^a f(x) \, dx = f(a), \quad f(1) = 1, \quad f(16) = \frac{1}{8}, \quad \text{then } 16 - f^{-1}\left( \frac{1}{16} \right) \text{ is equal to:}\]