The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these

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

Great question! When we arrange letters, we assume they are distinct. However, if some letters are identical, swapping them doesn't create a new, unique word. Since we have two $U$s, any arrangement where we swap these two $U$s is actually the same word. Dividing by $2!$ (which is $2$) removes these duplicate counts, ensuring we only count unique permutations.

Q: Why do we stop counting once we reach the target word?

The rank of a word is defined by its position in the dictionary. We are essentially counting how many words come *before* our target word. Once we reach $UDAYPUR$, we have accounted for all words that precede it. The rank is simply the total count of those words plus one (for the word itself).

Q: What if the word had more repeated letters, like 'BANANA'?

The logic remains the same! If you had more repetitions, you would just divide by the factorial of the count of *each* repeated letter. For 'BANANA', you have three $A$s and two $N$s. So, for any permutation calculation, you would divide by $3! \times 2!$. The 'fix and count' method is universal for any word!

Question

The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is

Options:

(A) 1579

(B) 1581

(C) 1578

(D) 1580

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the rank of the word $UDAYPUR$, we first identify the unique letters and their frequencies. The letters are $U, D, A, Y, P, U, R$.

Arranging these letters in alphabetical order, we get: $A, D, P, R, U, U, Y$. Note that the letter $U$ appears twice.

When calculating permutations, we must account for this repetition by dividing by $2!$ to avoid overcounting.

### The 'Before U' Marathon

In a dictionary, words are ordered alphabetically. Any word starting with $A, D, P,$ or $R$ will appear before any word starting with $U$.

For each of these $4$ starting letters, we have $6$ remaining positions to fill. The number of ways to arrange the remaining letters is:

$$
\frac{6!}{2!} = \frac{720}{2} = 360
$$

Since there are $4$ such starting letters, the total number of words appearing before those starting with $U$ is:

$$
4 	imes 360 = 1440
$$

### The 'U' Deep Dive

Now, we consider words starting with $U$. Since our target word $UDAYPUR$ starts with $U$, we lock it in the first position.

We look for words starting with $UA$, as $A$ is the only letter alphabetically preceding $D$. With $U$ and $A$ fixed, we have $5$ letters remaining ($D, P, R, U, Y$).

The number of arrangements for these is $5! = 120$. Adding this to our previous total:

$$
1440 + 120 = 1560
$$

### The Final Stretch: UDA...

Next, we fix $U$ and $D$. The target word continues with $A$, so we look for words starting with $UDA$.

The remaining letters are $P, R, U, Y$. We must count words starting with $UDAP, UDAR,$ and $UDAU$, as these precede $UDAY$.

For each of these $3$ prefixes, we have $3! = 6$ arrangements. This adds:

$$
3 	imes 6 = 18 	ext{ words}
$$

Our running total is now $1560 + 18 = 1578$.

### The Last Few Steps

We have fixed $UDAY$. The next letter in our target is $P$. The remaining letters are $P, R, U$.

Alphabetically, $P$ is the first choice, which matches our target. We fix $UDAYP$. Now we have $R$ and $U$ left.

The dictionary order for the remaining letters is $R$ then $U$. Thus, the next word is $UDAYPRU$ (rank $1579$).

The word immediately following is $UDAYPUR$. Therefore, the rank of $UDAYPUR$ is **$1580$**.

Animated Solution for Mathematics - Permutations and Combinations: The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is

Select Answer:

Visualized Solution

Alphabetical Arrangement

Words Starting with $A, D, P, R$

Fixing the First Letter $U$

Words Starting with $U A$

Fixing $U D$ and $U D A$

Words before $U D A Y$

Fixing $U D A Y$ and $U D A Y P$

Final Steps to $U D A Y P U R$

The Sigma Insight: Linear Permutations

Analyzing the Setup

The 'Before U' Marathon

The 'U' Deep Dive

The Final Stretch

The Last Few Steps

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