JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 18)

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:
OBBJH
HBBJO
OBBHJ
JBBOH

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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