The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of

Q: Why can't I just use the permutation formula $^{11}P_4$? Isn't that the standard way to arrange 4 letters from 11?

That is the most common trap in combinatorics! The formula $^{n}P_r$ assumes all $n$ objects are distinct. In 'EXAMINATION', the letters are not distinct—we have repeating A's, I's, and N's. If you treat them as distinct, you will overcount significantly because swapping two identical 'A's doesn't create a new word. We must group them to handle the repetitions correctly.

Q: In Case II, why do we use $^7C_2$ instead of $^8C_2$?

Great observation! We have 8 distinct types of letters available (A, I, N, E, X, M, T, O). In Case II, we have already 'used up' one type to form our pair (e.g., we picked the pair of A's). Since we need the remaining 2 letters to be distinct from the pair AND distinct from each other, we must choose from the remaining 7 types. If we chose from 8, we might accidentally pick the same letter we used for the pair, which would violate the condition of having exactly one pair.

Q: How do I know if I've missed a case? Is there a Case IV?

To check for missing cases, look at the structure of a 4-letter word. It can have: (1) 4 distinct letters, (2) 2 same + 2 distinct, (3) 2 same + 2 same, (4) 3 same + 1 distinct, or (5) 4 same. Looking at our inventory, we have no letter appearing 3 or 4 times. Therefore, cases (4) and (5) are impossible. You have covered all possibilities!

Question

The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word EXAMINATION is

Sigma Percentile · Accepted Answer

### Analyzing the Anatomy of the Word

Welcome, future engineer. Today, we are not just solving a permutation problem; we are performing an autopsy on the word 'EXAMINATION'. In the world of JEE Advanced, combinatorics is about the art of counting without double-counting.

When you see a word like 'EXAMINATION', your first instinct should be to categorize the chaos. Imagine you are standing in front of a whiteboard with the eleven letters: $E, X, A, M, I, N, A, T, I, O, N$.

Let us perform our inventory:
- We have pairs: $(A, A)$, $(I, I)$, and $(N, N)$. That is $3$ pairs.
- We have singles: $E, X, M, T, O$. That is $5$ single letters.

This gives us a total of $3 + 5 = 8$ distinct types of letters. We are going to build 4-letter words from this universe while respecting the constraints of repetition.

### The Strategy: Divide and Conquer

When we form a 4-letter word, the composition can vary. It could be all different, it could have one pair, or it could have two pairs.

These scenarios are mutually exclusive. We will calculate the number of ways for each case and sum them up using the Addition Principle.

### Case I: The Diversity of the Distinct

What if we choose 4 letters that are all completely different? We have $8$ distinct types of letters available and we need to pick $4$ and arrange them.

We use the permutation formula $^8P_4$:

$$^8P_4 = 8 	imes 7 	imes 6 	imes 5 = 1680$$

This represents selecting $4$ distinct letters from $8$ available types and arranging them in $4!$ ways. There are $1680$ such words.

### Case II: The Subtle Balance

Now, let us consider words containing exactly one pair of identical letters and two other distinct letters (e.g., 'A A E X').

First, we select the pair from the $3$ available pairs: $^3C_1$ ways. Next, we choose $2$ more letters from the remaining $7$ types: $^7C_2$ ways.

Now, we arrange these $4$ letters. Since $2$ are identical, we divide by $2!$ to remove overcounted permutations:

$$Number \ of \ ways = ^3C_1 	imes ^7C_2 	imes \frac{4!}{2!} = 3 	imes 21 	imes 12 = 756$$

There are $756$ words where exactly one pair exists.

### Case III: The Rare Symmetry

Finally, what if our word has two pairs (e.g., 'A A N N')? We need to select $2$ pairs out of the $3$ available: $^3C_2$ ways.

The arrangement of these $4$ letters, with two sets of identical pairs, is given by the multinomial coefficient:

$$Number \ of \ ways = ^3C_2 	imes \frac{4!}{2!2!} = 3 	imes 6 = 18$$

Only $18$ words fit this highly symmetric pattern.

### The Final Synthesis

We have analyzed the word, partitioned the possibilities, and calculated each scenario with precision. The total number of 4-letter words is the sum of our three cases:

$$Total = 1680 + 756 + 18 = 2454$$

The final answer is **2454**. You have learned to look at a problem, break it down into its constituent parts, and build the solution step-by-step. Keep this systematic approach, and no problem will ever be too complex for you.

Animated Solution for Mathematics - Permutations and Combinations: The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word EXAMINATION is

Enter Numerical Value:

Visualized Solution

Analyzing the word EXAMINATION

Identifying Pairs and Singles

Total Distinct Letters

Case I: $4$ Distinct Letters

Calculating Case I

Case II: $2$ Same and $2$ Distinct

Calculating Case II

Case III: $2$ Same and $2$ Same

Calculating Case III

Total Number of Words

The Sigma Insight: Linear Permutations

Analyzing the Anatomy of the Word

The Strategy

Case I

Case II

Case III

The Final Synthesis

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