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^{ ext {th }} word is:

Question

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^{	ext {th }} word is:

Accepted Answer

Answer: A. OBBJH To find the 50^{	ext{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. First, calculate the total number of permutations of these letters: \frac{5!}{2!} = 60 Let's arrange the letters in alphabetical order first: B, B, H, J, O. We need to systematically count the words while following dictionary order: 1. Words starting with B: Next position letters: B, H, J, O Number of permutations: \frac{4!}{1!} = 24 words Since 24 words starting with 'B' exist and are less than 50, Move to next starting letter alphabetically. 2. Words starting with H: Next position letters: B, B, J, O Number of permutations: \frac{4!}{2!} = 12 words After 'B', tally becomes 24 (B-words) + 12 (H-words) = 36, needs more. 3. Words starting with J: Next position letters: B, B, H, O Number of permutations: \frac{4!}{2!} = 12 words Now tally is 36 + 12 = 48 words. Still 2 more to reach 50. 4. Words starting with O: Next position letters: B, B, H, J Number of permutations: \frac{4!}{2!} = 12 words Our answer must be here since 48 + 2 more = 50 total. First permutation: OBB H J Second permutation (50th word): OBB J H So, the 50^{	ext{th}} word is: OBBJH Thus, the correct answer is Option A: OBBJH.

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:

Answer

Explanation

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