60 words can be made using all the letters of the word $\mathrm{BHBJO}$, with or without meaning. If these words are written as in a dictionary, then the $50^{\text {th }}$ word is:

<p>To find the $50^{\text{th}}$ word formed by the letters of "BHBJO" as if listed in a dictionary, let's analyze the arrangement methodically. Given the letters are B, H, B, J, O, there are some repetitions with the letter B appearing twice.</p> <p>First, calculate the total number of permutations of these letters:</p> <p>$\frac{5!}{2!} = 60$</p> <p>Let's arrange the letters in alphabetical order first: B, B, H, J, O.</p> <p>We need to systematically count the words while following dictionary order:</p> <p>1. Words starting with B:</p> <ul> <li>Next position letters: B, H, J, O</li> <li>Number of permutations: $\frac{4!}{1!} = 24$ words</li> </ul> <p>Since 24 words starting with 'B' exist and are less than 50, Move to next starting letter alphabetically.</p> <p>2. Words starting with H:</p> <ul> <li>Next position letters: B, B, J, O</li> <li>Number of permutations: $\frac{4!}{2!} = 12$ words</li> </ul> <p>After 'B', tally becomes 24 (B-words) + 12 (H-words) = 36, needs more.</p> <p>3. Words starting with J:</p> <ul> <li>Next position letters: B, B, H, O</li> <li>Number of permutations: $\frac{4!}{2!} = 12$ words</li> </ul> <p>Now tally is 36 + 12 = 48 words.</p> <p>Still 2 more to reach 50.</p> <p>4. Words starting with O:</p> <ul> <li>Next position letters: B, B, H, J</li> <li>Number of permutations: $\frac{4!}{2!} = 12$ words</li> <li>Our answer must be here since 48 + 2 more = 50 total.</li> </ul> <p>First permutation: $OBB H J$</p> <p>Second permutation (50th word): $OBB J H$</p> <p>So, the $50^{\text{th}}$ word is: <b>OBBJH</b></p> <p>Thus, the correct answer is <strong>Option A: OBBJH</strong>.</p>