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

Q: Wait, why can't we just use ${}^{11}P_{4}$? It seems so much faster!

I completely understand the temptation! ${}^{11}P_{4}$ is the go-to formula when all items are distinct. However, the word 'EXAMINATION' is not a collection of unique items. It contains repetitions ($A, I, N$). If you use ${}^{11}P_{4}$, you are treating the two 'A's as if they were different entities (like $A_{1}$ and $A_{2}$). But in the word, swapping them doesn't create a new word. Using the permutation formula would lead to massive overcounting. We must respect the 'indistinguishability' of the repeated letters.

Q: Why do we divide by $2!$ in Case 2? I get the selection part, but the arrangement confuses me.

Think of it this way: if you have 4 distinct letters, there are $4!$ ways to arrange them. But in Case 2, you have one pair (say, two 'A's) and two distinct letters (say, 'E' and 'X'). If you arrange them as $A_{1}EXA_{2}$ and $A_{2}EXA_{1}$, they look exactly the same! The $2!$ in the denominator is our 'correction factor'—it effectively divides out all those duplicate arrangements that look identical because of the repeated letter.

Q: How do I know if I've missed a case? Is there a Case 4?

That is a brilliant question. The key is to look at the 'partition' of the number 4. We are forming a 4-letter word. The possible partitions of 4 are: (1+1+1+1), (2+1+1), and (2+2). These correspond exactly to our three cases: all distinct, one pair, and two pairs. Since there are no other ways to sum to 4 using integers, we can be mathematically certain that we have covered every single possibility. You have captured the entire universe of outcomes!

Question

The number of 4 letter words (with or without meaning) that can be made from the eleven letters of the word “EXAMINATION” is _________ .

Sigma Percentile · Accepted Answer

### The Architecture of 'EXAMINATION'

Welcome, student. Today, we are not just solving a permutation problem; we are performing an autopsy on a word. The word is 'EXAMINATION'.

At first glance, it looks like a simple string of eleven letters. But to a JEE aspirant, it is a treasure trove of combinatorial logic. When you see a word with repeating letters, your internal alarm should ring.

This is not a simple ${}^{n}P_{r}$ problem. This is a problem of structure, of cases, and of careful, methodical counting.

### Phase 1: The DNA of the Word

Before we touch a single formula, we must understand our raw materials. Let us break down 'EXAMINATION' into its constituent parts. We have $11$ letters in total.

Are they unique? Absolutely not. We have the letter $A$ appearing twice, the letter $I$ appearing twice, and the letter $N$ appearing twice. These are our 'pairs'.

What remains? We have $E, X, M, T, O$. These are our 'singles'.

In total, we have $8$ distinct types of letters ($A, I, N, E, X, M, T, O$) to work with. This distinction between 'types' and 'total letters' is the foundation of our entire solution. If you miss this, you miss the problem.

### Phase 2: The Strategy of Cases

Since we need to form a 4-letter word, we cannot use a one-size-fits-all formula. We must partition our task into mutually exclusive scenarios. Think of this as building a house: we have different blueprints depending on the materials we choose.

### Case I: The 'All Different' Scenario

Imagine we want to pick 4 letters, and we want them all to be unique. We have $8$ distinct types of letters available. We choose $4$ of them in $\binom{8}{4}$ ways.

Once we have these $4$ unique letters, how many ways can we arrange them? That is a simple $4!$.

So, our total for Case I is:

$$ \binom{8}{4} 	imes 4! = 70 	imes 24 = 1680 $$

This is our baseline.

### Case II: The 'One Pair' Scenario

Now, things get interesting. What if our 4-letter word contains exactly one pair of identical letters? We need to select one pair from our $3$ available pairs ($A, I, N$). That is $\binom{3}{1}$ ways.

Now, we need $2$ more letters to complete our 4-letter word. Since we have already used one type for our pair, we have $7$ types left. We choose $2$ from these $7$ in $\binom{7}{2}$ ways.

Now, the arrangement. We have $4$ letters, but $2$ are identical. Because of the pair, we must divide by $2!$ to remove the duplicates.

Putting it all together:

$$ \binom{3}{1} 	imes \binom{7}{2} 	imes \frac{4!}{2!} = 3 	imes 21 	imes 12 = 756 $$

### Case III: The 'Two Pairs' Scenario

Finally, the rarest case: our word consists of two pairs of identical letters. We have $3$ pairs available, and we need to choose $2$ of them. That is $\binom{3}{2}$ ways.

Now, we arrange these $4$ letters (two pairs). The arrangement formula for this is $\frac{4!}{2! 	imes 2!}$. This accounts for the redundancy of both pairs.

So, we have:

$$ \binom{3}{2} 	imes \frac{4!}{2! 	imes 2!} = 3 	imes \frac{24}{4} = 3 	imes 6 = 18 $$

### The Grand Summation

We have navigated the three possible worlds of this problem. We have the 'All Different' world ($1680$), the 'One Pair' world ($756$), and the 'Two Pairs' world ($18$).

The final step is the most satisfying part of the journey: we sum them up.

$$ 1680 + 756 + 18 = 2454 $$

There you have it. We didn't just calculate a number; we dissected the logic of the word. You have mastered the art of case-based counting.

Remember, in JEE, the problem isn't just about the calculation—it's about the strategy. Keep this clarity, and no permutation problem will ever intimidate you again. The final answer is **2454**.

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

Enter Numerical Value:

Visualized Solution

Analyze the word $E X A M I N A T I O N$

Identify Repeated Letters

Identify Single Letters

Defining the Cases

Case I: Selection & Arrangement

Case II: Selection

Case II: Arrangement

Case III: Selection

Case III: Arrangement

Final Summation

The Sigma Insight: Combinations and Selection

The Architecture of 'EXAMINATION'

Phase 1

Phase 2

Case I

Case II

Case III

The Grand Summation

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