The number of 6-letter words, with or without meaning, that can be formed using the letters of the w

Q: Why can't we just use the formula $5^6$ to find the total number of words?

That is a great question! The formula $5^6$ counts all possible 6-letter words where each of the 6 positions can be filled by any of the 5 letters. However, this includes words where letters appear only once or not at all. Our problem has a strict constraint: any letter that appears must appear at least twice. This means we must exclude any word that contains a letter appearing only once. Partitioning the problem into valid frequency sets is the only way to ensure we satisfy this constraint.

Q: In the $(4, 2)$ case, why do we multiply by $2!$?

Excellent observation. When we choose two letters, say M and A, we have two distinct frequencies to assign: 4 and 2. We could have a word with four Ms and two As, OR a word with four As and two Ms. These are distinct scenarios. The $2!$ accounts for the number of ways to assign the frequency '4' to one of the chosen letters and '2' to the other.

Q: Why don't we multiply by $2!$ in the $(3, 3)$ case?

Think about the assignment process. If we choose letters M and A, we need to assign the frequency '3' to one and '3' to the other. If we swap them, we still have three Ms and three As. Because the frequencies are identical, swapping the letters doesn't create a new assignment of frequencies. Therefore, multiplying by $2!$ would be redundant and would overcount the possibilities.

Question

The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with forming a six-letter word using the letters from the set $\{M, A, T, H, S\}$. The fundamental constraint is that every letter used in the word must appear at least twice.

This is a partitioning problem where we must express the total length of $6$ as a sum of integers, where each integer $n_i \ge 2$. The possible partitions are:

1.  A single letter appearing $6$ times: $(6)$.
2.  Two letters, one appearing $4$ times and the other $2$ times: $(4, 2)$.
3.  Two letters, each appearing $3$ times: $(3, 3)$.
4.  Three letters, each appearing $2$ times: $(2, 2, 2)$.

These cases are mutually exclusive and collectively exhaustive.

### The Calculation

We now calculate the number of arrangements for each case systematically.

**Case 1: The $(6)$ pattern**
We choose $1$ letter from $5$ in $^5C_1 = 5$ ways. The number of arrangements for $6$ identical letters is:
$$\frac{6!}{6!} = 1$$
Total for Case 1: $5 	imes 1 = 5$.

**Case 2: The $(4, 2)$ pattern**
We choose $2$ letters from $5$ in $^5C_2 = 10$ ways. Since the frequencies ($4$ and $2$) are distinct, we have $2! = 2$ ways to assign these frequencies to the chosen letters. The number of arrangements is:
$$\frac{6!}{4!2!} = 15$$
Total for Case 2: $10 	imes 2 	imes 15 = 300$.

**Case 3: The $(3, 3)$ pattern**
We choose $2$ letters from $5$ in $^5C_2 = 10$ ways. Since the frequencies are identical, no further assignment is required. The number of arrangements is:
$$\frac{6!}{3!3!} = 20$$
Total for Case 3: $10 	imes 20 = 200$.

**Case 4: The $(2, 2, 2)$ pattern**
We choose $3$ letters from $5$ in $^5C_3 = 10$ ways. Again, the frequencies are identical, so no assignment is needed. The number of arrangements is:
$$\frac{6!}{2!2!2!} = \frac{720}{8} = 90$$
Total for Case 4: $10 	imes 90 = 900$.

### Final Calculation

To find the total number of valid words, we sum the results from all four cases:

$$5 + 300 + 200 + 900 = 1405$$

The total number of ways to form the six-letter word under the given constraints is **1405**.

Animated Solution for Mathematics - Permutations and Combinations: The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is

Enter Numerical Value:

Visualized Solution

Analyzing the Word MATHS

Identifying Frequency Patterns

Case 1: Pattern $(6)$

Case 2: Pattern $(4, 2)$

Case 3: Pattern $(3, 3)$

Case 4: Pattern $(2, 2, 2)$

Final Summation

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Calculation

Final Calculation

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