60 words can be made using all the letters of the word BHBJO, with or without meaning. If these word

Q: Wait, why do we divide by $2!$ in the permutations?

Great question! When we calculate permutations, we assume all items are distinct. However, if we have two 'B's, swapping them doesn't create a new, unique word. Dividing by $2!$ removes these redundant, identical arrangements, ensuring we only count unique words.

Q: How do I know when to stop counting?

You stop when the cumulative sum of your blocks reaches or exceeds your target rank. In this case, we counted blocks of 24, 12, and 12, reaching 48. Since 48 is just shy of 50, we know the 50th word must lie within the next block, which starts with 'O'.

Q: Is there a faster way than manual counting?

For small words like this, systematic counting is the most reliable method. It prevents errors. While there are formulas for rank, they are essentially just a formal way of doing exactly what we did: counting how many words come before the target.

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 :

Options:

(A) JBBOH

(B) OBBJH

(C) OBBHJ

(D) HBBJO

Sigma Percentile · Accepted Answer

### The Art of Systematic Counting

Welcome, future engineer! Today, we are diving into the elegant world of combinatorics. Dictionary rank problems are not just about memorizing formulas; they are about developing a systematic, logical mindset.

Imagine you are a librarian organizing a very strange, five-letter dictionary. We have the letters B, B, H, J, and O. Our goal is to find the $50^{th}$ word.

### Phase 1: The Alphabetical Foundation

Before we count, we must organize. The dictionary is a place of order. If we sort our letters alphabetically, we get: B, B, H, J, O.

This is our roadmap. Every word in our dictionary will follow this sequence. We will count how many words start with B, then H, then J, and finally O, until we hit our target of 50.

### Phase 2: The Systematic March

Let's start with the first letter, B. If we fix B at the first position, we have four slots left. The remaining letters are B, H, J, and O.

Since these four are distinct, the number of ways to arrange them is:

$$4! = 24$$

So, there are 24 words starting with B. Our cumulative count is 24. We need to reach 50, so we keep going.

Next is H. If we fix H at the first position, we have four slots left. The remaining letters are B, B, J, and O.

Here is the catch: the letter B repeats twice. The number of arrangements is:

$$\frac{4!}{2!} = 12$$

Adding this to our previous count, $24 + 12 = 36$. Still not at 50.

Next is J. If we fix J at the first position, we have four slots left. The remaining letters are B, B, H, and O.

Again, B repeats twice. The number of arrangements is:

$$\frac{4!}{2!} = 12$$

Adding this to our previous count, $36 + 12 = 48$. We are so close!

### Phase 3: The Final Leap

We have accounted for 48 words. The $49^{th}$ word must start with the next letter, O.

To find the very first word starting with O, we arrange the remaining letters (B, B, H, J) in strict alphabetical order. That gives us **OBBHJ**. This is our $49^{th}$ word.

Now, for the $50^{th}$ word, we simply need the next permutation after **OBBHJ**. We look at the suffix HJ and swap them to get JH.

Thus, the $50^{th}$ word is **OBBJH**.

See how beautiful that is? By breaking the problem into manageable blocks, we turned a daunting task into a simple, logical journey. Keep practicing this systematic approach, and you will master combinatorics in no time!

Animated Solution for Mathematics - Permutations and Combinations: 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 $5 0^{t h}$ word is :

Select Answer:

Visualized Solution

Analyze the Word $B H B J O$

Alphabetical Sorting

Words Starting with $B$

Words Starting with $H$

Words Starting with $J$

The $4 9^{t h}$ Word

The $5 0^{t h}$ Word

The Sigma Insight: Linear Permutations

The Art of Systematic Counting

Phase 1

Phase 2

Phase 3

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