JEE Main 2023 (01 February Shift 1)

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Animated Solution for Mathematics - Permutations and Combinations: The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is ____.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Think of the word 'ASSASSINATION' not as a static string, but as a collection of individual characters. Before you start counting, identify the 'vowel group' and the 'consonant group'. The constraint is that the vowels must act as a single, inseparable entity.

Hint 2

Hint 2: When you need to keep items together, the 'Tie Method' is your best friend. Imagine tying all the vowels into a single bag. Now, how many items are you actually arranging? Remember, the items inside the bag can also rearrange themselves, and you must account for the identical letters in both the consonant group and the vowel group.

Hint 3

Hint 3: Your final calculation will be the product of two distinct permutations: the arrangement of the 'units' (consonants plus the vowel bag) and the internal arrangement of the vowels within that bag. Use the formula for permutations of multisets:

P = \frac{n !}{n _{1} ! n _{2} ! ... n _{k} !}

Visualized Solution

Analyze the Word

Word: ASSASSINATION
Total letters: $13$

The Core Condition

Condition: All vowels must occur together.

Segregating Letters

Consonants: $S, S, S, S, N, N, T$ (Total $7$ )
Vowels: $A, A, A, I, I, O$ (Total $6$ )

The Tie Method

To keep vowels together, we tie them into a single block.
Treat the $6$ vowels as one unit.

Counting Total Units

$7$ Consonants + $1$ Vowel Block = $8$ Units

Arranging the Units

We need to arrange these $8$ units.
Wait, are all units distinct?

Handling Repetitions

Among the consonants, we have identical letters:
$4$ S's
$2$ N's

Permutation of Units

Ways to arrange $8$ units = $\frac{8 !}{4 ! 2 !}$

Internal Arrangement

The vowels inside the block can also shuffle among themselves!

Vowel Repetitions

Inside the block, we have $6$ vowels:
$3$ A's
$2$ I's

Permutation of Vowels

Ways to arrange $6$ vowels = $\frac{6 !}{3 ! 2 !}$

Total Combinations

Total words = (Arrangement of Units) $\times$ (Internal Arrangement)

The Final Answer

Total words = $\frac{8 !}{4 ! 2 !} \times \frac{6 !}{3 ! 2 !}$
$= 840 \times 60$
$= 50400$

00:00 / 00:00

The Sigma Insight: Linear Permutations

Solution Diagram

Analyzing the Setup

The word 'ASSASSINATION' consists of 13 letters in total. To solve this, we must first categorize the letters into two distinct groups: consonants and vowels.

The consonants are:

S, S, S, S, N, N, T

. This gives us a total of 7 consonants.

The vowels are:

A, A, A, I, I, O

. This gives us a total of 6 vowels.

The Tie Method

When the problem dictates that specific items must remain together, we treat them as a single entity. By placing all 6 vowels into a single "vowel-bag," we reduce our total number of items to arrange.

We now have 7 consonants and 1 "vowel-bag," resulting in

7 + 1 = 8

units to arrange.

However, these 8 units are not all unique. We must account for the 4 identical 'S's and 2 identical 'N's. The number of ways to arrange these 8 units is given by:

Arrangements of Units = \frac{8 !}{4 ! 2 !}

Calculating this, we find:

\frac{40320}{24 \times 2} = 840

The Internal Dance

The vowels inside the "vowel-bag" are not static; they can be arranged in various ways among themselves. We must calculate the permutations of the set

{A, A, A, I, I, O}

Because there are three 'A's and two 'I's, we use the formula for permutations of a multiset:

Internal Arrangements = \frac{6 !}{3 ! 2 !}

Performing the calculation:

\frac{720}{6 \times 2} = 60

The Grand Finale

According to the Fundamental Counting Principle, the total number of distinct arrangements is the product of the external arrangements and the internal arrangements.

Total Words = 840 \times 60

The final result is 50,400.

By applying the 'Tie Method' and accounting for the repetitions within the sets, we have successfully brought order to the chaos of the 13-letter string.

Common Traps & FAQs

Wait, why do we treat the vowel block as a single unit? Doesn't that change the total number of letters?

That is a brilliant observation! By treating the vowels as one unit, we are essentially creating a 'super-letter'. We aren't changing the total number of letters in the word; we are simply imposing a constraint on their relative positions. By gluing them together, we ensure they never drift apart, effectively reducing the number of items to arrange from 13 down to 8.

Why do we divide by

4!

and

2!

in the first part? I thought we just arrange the units.

Remember, when we arrange items, we assume they are distinct. However, in 'ASSASSINATION', the letters 'S' and 'N' are identical. If you swap one 'S' with another 'S', the word looks exactly the same. Dividing by the factorial of the count of identical items is the mathematical way of 'removing' these redundant, invisible swaps.

Do I need to worry about the internal arrangement of the consonants too?

You already handled that! When we arranged the 8 units (7 consonants + 1 vowel block), we accounted for the repetitions of 'S' and 'N' within that set. The internal arrangement of the consonants is already baked into the

\frac{8 !}{4 ! 2 !}

term.

Is it possible to solve this by subtracting the cases where vowels are separated?

Technically, yes, but that would be a nightmare! You would have to use the Principle of Inclusion-Exclusion to subtract cases where vowels are apart, which is significantly more complex than the 'Tie Method'. Always choose the path of least resistance in competitive exams.

What if the question asked for vowels to be together AND consonants to be together?

Then you would have two blocks: one for all vowels and one for all consonants. You would arrange these two blocks (

2!

ways) and then multiply by the internal permutations of the vowel block and the consonant block separately. It is just an extension of the same logic!

Animated Solution for Mathematics - Permutations and Combinations: The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is ____.

Enter Numerical Value:

Visualized Solution

Analyze the Word

The Core Condition

Segregating Letters

The Tie Method

Counting Total Units

Arranging the Units

Handling Repetitions

Permutation of Units

Internal Arrangement

Vowel Repetitions

Permutation of Vowels

Total Combinations

The Final Answer

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Tie Method

The Internal Dance

The Grand Finale

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