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 50th word is :

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

Answer 1

Answer: (C)

OBBJH

Answer 2

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.