Question

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

• A. OBBHJ
• B. HBBJO
• C. OBBJH
• D. JBBOH

Answer 1

The Correct Option is C

To find the 50th word in the dictionary order of permutations of the letters in BHBJO, we proceed as follows:

Step 1: Arrange the letters alphabetically

The letters in the word BHBJO, arranged in alphabetical order, are:

\[ B, B, H, J, O. \]

Step 2: Total number of permutations

The total number of permutations of these letters is:

\[ \frac{5!}{2!} = \frac{120}{2} = 60 \quad (\text{since there are 2 repeated } B's). \]

Step 3: Determine the block sizes

Fix the first letter in alphabetical order and calculate the number of permutations for each block.

Case 1: First letter \(B\)

If the first letter is \(B\), the remaining letters are \(B, H, J, O\). The number of permutations of these is:

\[ \frac{4!}{2!} = \frac{24}{2} = 12. \]

Thus, the first 12 words start with \(B\).

Case 2: First letter \(H\)

If the first letter is \(H\), the remaining letters are \(B, B, J, O\). The number of permutations of these is:

\[ \frac{4!}{2!} = \frac{24}{2} = 12. \]

Thus, the next 12 words (from 13 to 24) start with \(H\).

Case 3: First letter \(J\)

If the first letter is \(J\), the remaining letters are \(B, B, H, O\). The number of permutations of these is:

\[ \frac{4!}{2!} = \frac{24}{2} = 12. \]

Thus, the next 12 words (from 25 to 36) start with \(J\).

Case 4: First letter \(O\)

If the first letter is \(O\), the remaining letters are \(B, B, H, J\). The number of permutations of these is:

\[ \frac{4!}{2!} = \frac{24}{2} = 12. \]

Thus, the next 12 words (from 37 to 48) start with \(O\).

Step 4: Focus on the 49th to 60th words

The 49th to 60th words start with \(OB\), since the first letter is \(O\) and the second letter must now be \(B\). The remaining letters to permute are \(B, H, J\). These permutations are:

\[ OBBHJ, OBBJH, OBHBJ, OBJHB, OBJBH, OBJHB. \]

The second word in this list is the 50th word: OBBJH.

Answer 2

Step 1: Write down the letters in alphabetical order.
The letters of the word BHBJO are: A–not present, so arrange existing letters alphabetically → B, B, H, J, O.

Step 2: Total number of words possible.
Total letters = 5, with two B’s repeated.
Number of permutations = 5! / 2! = 60.

Step 3: Find how many words start with each letter.
• Starting with B: remaining letters = B, H, J, O → 4! / 1! = 12 words.
• Starting with H: remaining letters = B, B, J, O → 4! / 2! = 12 words.
• Starting with J: remaining letters = B, B, H, O → 4! / 2! = 12 words.
• Starting with O: remaining letters = B, B, H, J → 4! / 2! = 12 words.
• Starting with the second B already included (same as first B case).

Step 4: Word position groups.
1–12 : start with B
13–24 : start with H
25–36 : start with J
37–48 : start with O
49–60 : start with remaining letter (after O group completes).

The 49th word falls into the “starting with O” group (which covers 37–48), but since 50 is just after 48, it means the next group starts with the next lexicographic letter after O group—actually, within O group we have 37–48, and then 49–60 must start with the next letter sequence starting with O (recalculation below).

Wait: For 37–48 (O...), we had 12 words. So, after 48, we reach the next set, which starts with OB (as O with B next). Let's find the arrangement inside O group precisely.

Step 5: Expand O group (O + remaining B, B, H, J).
Possible arrangements of B, B, H, J = 4! / 2! = 12 words.
These cover positions 49–60.

Let’s list them in order:
BBHJ, BBJH, BH BJ (we can systematically find the 50th).
But for O + (BBHJ) type words:
1st (49th overall): OBBHJ
2nd (50th overall): OBBJH

Step 6: Final Answer.
The 50th word = OBBJH.

Final Answer: OBBJH