JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 18)
Explanation
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.
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 $$50^{\text{th}}$$ word is: OBBJH
Thus, the correct answer is Option A: OBBJH.
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