Ivan wants to create a word using the letters from ' \( A B C^{\prime} \) only. He may use the same letters as much as he wants. How many 6 -letter words can he make such that the letters ' \( B \) ' appear at most 4 times?

To solve the problem, we need to count the number of 6-letter words that can be formed using the letters \( A, B, C' \) with the condition that the letter \( B \) appears at most 4 times. We will break this down into cases based on the number of times \( B \) appears in the word. ### Case 1: \( B \) appears 0 times In this case, all 6 letters must be either \( A \) or \( C' \). Therefore, we have 2 choices for each of the 6 positions. \[ \text{Number of words} = 2^6 = 64 \] ### Case 2: \( B \) appears 1 time We can choose 1 position for \( B \) in \( \binom{6}{1} \) ways. The remaining 5 positions can be filled with either \( A \) or \( C' \). \[ \text{Number of ways to choose position for } B = \binom{6}{1} = 6 \] \[ \text{Number of ways to fill remaining 5 positions} = 2^5 = 32 \] \[ \text{Total for this case} = 6 \times 32 = 192 \] ### Case 3: \( B \) appears 2 times We can choose 2 positions for \( B \) in \( \binom{6}{2} \) ways. The remaining 4 positions can be filled with either \( A \) or \( C' \). \[ \text{Number of ways to choose positions for } B = \binom{6}{2} = 15 \] \[ \text{Number of ways to fill remaining 4 positions} = 2^4 = 16 \] \[ \text{Total for this case} = 15 \times 16 = 240 \] ### Case 4: \( B \) appears 3 times We can choose 3 positions for \( B \) in \( \binom{6}{3} \) ways. The remaining 3 positions can be filled with either \( A \) or \( C' \). \[ \text{Number of ways to choose positions for } B = \binom{6}{3} = 20 \] \[ \text{Number of ways to fill remaining 3 positions} = 2^3 = 8 \] \[ \text{Total for this case} = 20 \times 8 = 160 \] ### Case 5: \( B \) appears 4 times We can choose 4 positions for \( B \) in \( \binom{6}{4} \) ways. The remaining 2 positions can be filled with either \( A \) or \( C' \). \[ \text{Number of ways to choose positions for } B = \binom{6}{4} = 15 \] \[ \text{Number of ways to fill remaining 2 positions} = 2^2 = 4 \] \[ \text{Total for this case} = 15 \times 4 = 60 \] ### Total Count Now, we sum the totals from all cases: \[ \text{Total} = 64 + 192 + 240 + 160 + 60 = 716 \] Thus, the total number of 6-letter words that can be formed such that the letter \( B \) appears at most 4 times is \[ \boxed{716} \]