The number of four letter words that can be formed using the letters of the word BARRACK is :- (A) 2

Q: Why can't I just use the permutation formula $^7P_4$?

That formula assumes all $7$ letters are distinct. Because 'BARRACK' has repeating 'A's and 'R's, $^7P_4$ would overcount significantly. We must group by cases to account for the identical items.

Q: In Case 2, why do we select the pair first?

Because the pair is a special entity. By selecting the pair first, we ensure we don't accidentally pick the same letter twice in the 'different' category. It simplifies the selection process.

Q: How do I know if I've missed a case?

A systematic approach is your best friend. By categorizing based on the number of identical letters (e.g., $2$ alike, $2$ alike; $2$ alike, $2$ different; all different), you cover all possibilities without overlap. If you follow this partition, you won't miss any.

Question

The number of four letter words that can be formed using the letters of the word BARRACK is :-

Options:

(A) 270

(B) 120

(C) 264

(D) 144

Sigma Percentile · Accepted Answer

### Analyzing the Anatomy of BARRACK

The word 'BARRACK' consists of $7$ letters. Upon inspection, we identify the following frequency distribution: two 'A's, two 'R's, one 'B', one 'C', and one 'K'.

This implies we have $5$ distinct types of letters: $\{A, R, B, C, K\}$. To form a $4$-letter word, we must account for these repetitions systematically to avoid errors.

### The Three Pillars of the Solution

To solve this, we partition the problem into three mutually exclusive cases based on the selection of letters:

*   **Case 1:** Choosing two pairs of letters (e.g., $A, A, R, R$).
*   **Case 2:** Choosing one pair and two distinct letters (e.g., $A, A, B, C$).
*   **Case 3:** Choosing four completely different letters (e.g., $A, R, B, C$).

### Diving into the Cases

**Case 1: Two pairs**
We have two pairs available: $\{A, A\}$ and $\{R, R\}$. We select both pairs in $^2C_2 = 1$ way.

The number of arrangements for these four letters is:
$$
\frac{4!}{2! 	imes 2!} = 6
$$
Thus, Case 1 yields $6$ words.

**Case 2: One pair and two different letters**
We choose one pair from the $2$ available types in $^2C_1 = 2$ ways. We then select two additional distinct letters from the remaining $4$ types in $^4C_2 = 6$ ways.

The total number of selections is $2 	imes 6 = 12$. The number of arrangements for each selection (where two letters are identical) is:
$$
\frac{4!}{2!} = 12
$$
Therefore, Case 2 yields $12 	imes 12 = 144$ words.

**Case 3: All four different**
We have $5$ distinct types of letters available: $\{A, R, B, C, K\}$. We select $4$ distinct types in $^5C_4 = 5$ ways.

The number of arrangements for $4$ distinct letters is $4! = 24$. Thus, Case 3 yields:
$$
5 	imes 24 = 120 	ext{ words}
$$

### The Final Synthesis

To find the total number of $4$-letter words, we sum the results from our three pillars:
$$
6 + 144 + 120 = 270
$$

The total number of $4$-letter words that can be formed from the letters of the word 'BARRACK' is **$270$**.

Animated Solution for Mathematics - Permutations and Combinations: The number of four letter words that can be formed using the letters of the word BARRACK is :-

Select Answer:

Visualized Solution

Analyzing the Word

Frequency of Letters

Distinct Letter Types

Defining the Cases

Case 1: $2$ Alike, $2$ Alike (Selection)

Case 1: Arrangement

Case 2: $2$ Alike, $2$ Different (Selection)

Case 2: Selecting Different Letters

Case 2: Arrangement

Case 3: All $4$ Different (Selection)

Case 3: Arrangement

Total Number of Words

The Sigma Insight: Linear Permutations

Analyzing the Anatomy of BARRACK

The Three Pillars of the Solution

Diving into the Cases

The Final Synthesis

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