JEE Main 2023 (06 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Permutations and Combinations: All the letters of the word PUBLIC are written in all possible orders and these words are written as in a dictionary with serial numbers. Then the serial number of the word PUBLIC is

Q: Wait, why do we add 1 at the very end of the calculation?

That is a fantastic question! The calculation $480 + 96 + 4 + 1$ gives us the number of words that appear *before* the word PUBLIC in the dictionary. Since we want the *rank* (or serial number) of the word PUBLIC itself, we must include it in the count. Think of it like a queue: if 581 people are standing in front of you, you are the 582nd person. Hence, we add 1.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. To find the dictionary rank, you must first establish the alphabetical order of the letters in the word PUBLIC. List them out: B, C, I, L, P, U. Visualize the dictionary as a sequence of words where we systematically count how many words appear before our target word by fixing one position at a time.

Hint 2

Hint 2: The Arsenal. Use the fundamental principle of counting permutations. For any position you fix, the number of words that can be formed with the remaining letters is given by the factorial of the number of remaining positions. Remember to skip all letters that come alphabetically before the letter in your target word at that specific position.

Hint 3

Hint 3: The Execution. Start by counting words that begin with B, C, I, and L. The number of such words is

4 \times 5!

. Then, move to the second letter, P, and count words starting with PB, PC, PI, and PL. Continue this process for each subsequent letter until you reach the word PUBLIC itself, and don't forget to add

1

at the very end for the word itself.

Visualized Solution

Alphabetical Arrangement

Word: PUBLIC
Letters: P, U, B, L, I, C
Alphabetical Order: B, C, I, L, P, U

Words Before 'P'

Target word starts with P.
We must skip all words starting with B, C, I, L.
Number of such starting letters = $4$ .

Counting Words Before 'P'

For each starting letter (B, C, I, L), there are $5$ remaining positions.
Arrangements per letter = $5! = 120$ .
Total words skipped = $4 \times 5! = 480$ .

Fixing 'P' and Moving to 'U'

Now we are in the P section.
Target word is PU...
Remaining letters: B, C, I, L, U.
We must skip words starting with PB, PC, PI, PL.

Counting Words Before 'PU'

Number of prefixes to skip = $4$ (PB, PC, PI, PL).
Remaining positions for each = $4$ .
Total words skipped = $4 \times 4! = 96$ .

Fixing 'PU' and Finding 'B'

Now we are in the PU section.
Target word is PUB...
Remaining letters: B, C, I, L.
B is the first available letter! No words to skip.

Fixing 'PUB' and Moving to 'L'

Now we are in the PUB section.
Target word is PUBL...
Remaining letters: C, I, L.
We must skip words starting with PUBC, PUBI.

Counting Words Before 'PUBL'

Number of prefixes to skip = $2$ (PUBC, PUBI).
Remaining positions for each = $2$ .
Total words skipped = $2 \times 2! = 4$ .

Fixing 'PUBL' and Moving to 'I'

Now we are in the PUBL section.
Target word is PUBLI...
Remaining letters: C, I.
We must skip words starting with PUBLC.

Counting Words Before 'PUBLI'

Number of prefixes to skip = $1$ (PUBLC).
Remaining positions = $1$ .
Total words skipped = $1 \times 1! = 1$ .

The Word 'PUBLIC'

Now we are in the PUBLI section.
Remaining letter: C.
The very next word is PUBLIC.
Count for this word = $1$ .

Summing Up the Rank

Rank of PUBLIC = Sum of all skipped words + $1$
Rank = $480 + 96 + 4 + 1 + 1$
Rank = $582$

00:00 / 00:00

The Sigma Insight: Linear Permutations

Solution Diagram

Analyzing the Setup

To find the rank of the word PUBLIC, we first determine the alphabetical order of its constituent letters: B, C, I, L, P, U.

This sorted sequence serves as our reference for all permutations. The total number of letters is

6

, and all letters are distinct.

The First Leap

Bypassing the Early Sections

We aim to reach words starting with P. In a dictionary, all words starting with B, C, I, or L appear before those starting with P.

There are

4

such starting letters. For each, the remaining

5

positions can be filled in

5!

ways.

4 \times 5! = 4 \times 120 = 480

We have now bypassed

480

words.

The P-Section

Precision Navigation

Now we are in the P section. Our target word is PU.... Before reaching PU, we must pass through all words starting with PB, PC, PI, and PL.

There are

4

such prefixes. For each, the remaining

4

positions can be filled in

4!

ways.

4 \times 4! = 4 \times 24 = 96

Adding this to our previous tally, we get

480 + 96 = 576

The Final Stretch

Locking in the Letters

We are now in the PU section. The third letter of our target is B. Since B is the first available letter in our remaining set

{B, C, I, L}

, we skip

0

words.

Next, we look at the fourth letter, L. In the remaining set

{C, I, L}

, the letters C and I precede L. We must skip words starting with PUBC and PUBI.

Each of these

2

prefixes has

2!

arrangements:

2 \times 2! = 2 \times 2 = 4

Our running total is now

576 + 4 = 580

Final Calculation

The fifth letter is I. In the remaining set

{C, I}

, the letter C comes before I. We must skip the word starting with PUBLC.

This accounts for

1 \times 1! = 1

word. Our count becomes

580 + 1 = 581

Finally, the only remaining letter is C, which completes the word PUBLIC. Since this is the very next word in the sequence, we add

1

to our total.

The final rank of the word PUBLIC is 582.

Common Traps & FAQs

Wait, why do we add 1 at the very end of the calculation?

That is a fantastic question! The calculation

480 + 96 + 4 + 1

gives us the number of words that appear *before* the word PUBLIC in the dictionary. Since we want the *rank* (or serial number) of the word PUBLIC itself, we must include it in the count. Think of it like a queue: if 581 people are standing in front of you, you are the 582nd person. Hence, we add 1.

What if the word had repeating letters? Would the factorial method still work?

Great intuition! If there were repeating letters, the simple factorial method would overcount. You would need to use the formula for permutations of a multiset:

\frac{n !}{n _{1} ! n _{2} ! \dots n _{k} !}

, where

n

is the total number of letters and

n_{1}, n_{2}, \dots

are the frequencies of the repeating letters. For PUBLIC, all letters are distinct, so

n!

works perfectly!

Why can't we just use a simpler method to find the rank?

The dictionary rank problem is a classic exercise in systematic counting. While there are shortcuts for specific types of problems, the 'fixing and skipping' method is the most robust and least prone to errors. It forces you to respect the lexicographical structure of the dictionary, which is the core concept being tested in JEE Advanced.

Animated Solution for Mathematics - Permutations and Combinations: All the letters of the word PUBLIC are written in all possible orders and these words are written as in a dictionary with serial numbers. Then the serial number of the word PUBLIC is

Select Answer:

Visualized Solution

Alphabetical Arrangement

Words Before 'P'

Counting Words Before 'P'

Fixing 'P' and Moving to 'U'

Counting Words Before 'PU'

Fixing 'PU' and Finding 'B'

Fixing 'PUB' and Moving to 'L'

Counting Words Before 'PUBL'

Fixing 'PUBL' and Moving to 'I'

Counting Words Before 'PUBLI'

The Word 'PUBLIC'

Summing Up the Rank

The Sigma Insight: Linear Permutations

Analyzing the Setup

The First Leap

The P-Section

The Final Stretch

Final Calculation

Similar Questions

The letters of the word 'MANKIND' are written in all possible orders and arranged in serial order as in an English dictionary. Then the serial number of the word 'MANKIND' is ______.

The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is :

All the letters of the word "GTWENTY" are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word "GTWENTY" IS

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number

All words, with or without meaning, are made using all the letters of the word $M O N D A Y$ . These words are written as in a dictionary with serial numbers. The serial number of the word $M O N D A Y$ is

If the letters of the word MATHS are permuted and all possible words so formed are arranged as in a dictionary with serial numbers, then the serial number of the word THAMS is

All the arrangements, with or without meaning, of the word $F A R M E R$ are written excluding any word that has two $R$ appearing together. The arrangements are listed serially in the alphabetic order as in the English dictionary. Then the serial number of the word $F A R M E R$ in this list is .

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

If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is

If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is :