The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where th

Q: Wait, why can't we just use ${}^{11}C_5$ to solve this?

That is a common trap! The formula ${}^{11}C_5$ assumes all 11 letters are distinct, which they are not. Because the word MATHEMATICS contains repeating letters, selecting 'M' and 'A' is different from selecting 'M' and 'M'. We must account for these repetitions by grouping them into types.

Q: Why can't we have 3 pairs in our selection?

Great observation! We are only selecting 5 letters in total. If we were to select 3 pairs, that would require $3 \times 2 = 6$ letters, which exceeds our limit of 5. Therefore, the maximum number of pairs we can possibly include is 2.

Q: Why do we subtract types when choosing distinct letters in Case 2 and Case 3?

When we select a pair, we have already 'used' that letter type. For example, if we choose the pair 'M', we cannot choose 'M' again as a distinct letter because that would form another pair. Thus, we must exclude the types already chosen as pairs from the pool of available distinct letters.

Question

The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to :

Options:

(A) 179

(B) 177

(C) 181

(D) 175

Sigma Percentile · Accepted Answer

### Analyzing the Anatomy of MATHEMATICS

Welcome, fellow traveler, to the elegant world of combinatorics. Today, we are not just counting letters; we are dissecting the very structure of the word $MATHEMATICS$.

When you look at this word, do you see just a string of characters, or do you see a mathematical puzzle waiting to be solved? Let us peel back the layers.

### Phase 1: Deconstructing the Word

First, let us look at the word $MATHEMATICS$. It has a total of $11$ letters.

In combinatorics, identical items change the game entirely. We must group them:
*   We have two $M$'s, two $A$'s, and two $T$'s. These are our three types of pairs.
*   The remaining letters—$H, E, I, C, S$—are all unique.

In total, we have $8$ distinct types of letters to work with. This is our toolkit.

### Phase 2: The Strategy of Cases

Our goal is to select exactly $5$ letters. Because we have a mix of identical and distinct letters, we cannot use a single, simple formula.

Instead, we must be strategic. We will break this down into mutually exclusive cases based on how many pairs we include. This is the heart of the problem.

### Phase 3: Case 1 - The Distinct Path

What if all five letters we select are completely distinct? We have $8$ distinct types of letters available ($M, A, T, H, E, I, C, S$).

To choose $5$ distinct letters from $8$ types, we use the combination formula ${}^8C_5$:

$$
{}^8C_5 = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} = 56 	ext{ ways}
$$

This is our first foundation.

### Phase 4: Case 2 - The Single Pair Path

Now, let us introduce a pair. We select exactly $1$ pair (two identical letters) and $3$ distinct letters, which gives us $2 + 3 = 5$ letters in total.

First, we choose $1$ pair from the $3$ available pairs ($M, A, T$), which is ${}^3C_1 = 3$ ways. Now, we need $3$ more letters from the remaining $7$ distinct types (excluding the pair we just picked):

$$
{}^7C_3 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35 	ext{ ways}
$$

Multiplying these, we get $3 	imes 35 = 105$ ways.

### Phase 5: Case 3 - The Double Pair Path

Finally, what if we select $2$ pairs? That gives us $4$ letters, so we need exactly $1$ distinct letter to reach our total of $5$.

We choose $2$ pairs from the $3$ available, which is ${}^3C_2 = 3$ ways. We have used up $2$ types of letters, so we have $6$ distinct types remaining. We choose $1$ distinct letter from these $6$:

$$
{}^3C_2 	imes {}^6C_1 = 3 	imes 6 = 18 	ext{ ways}
$$

### The Grand Finale

Since these cases are mutually exclusive, we simply add them together:

$$
56 + 105 + 18 = 179
$$

There you have it! **One hundred and seventy-nine ways** to choose your letters.

It is not just about the calculation; it is about the logic of partitioning the chaos into order. You have mastered the structure of the problem. Keep this mindset, and no combinatorics problem will ever intimidate you again.

Animated Solution for Mathematics - Permutations and Combinations: The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to :

Select Answer:

Visualized Solution

Analyze the Word $M A T H E M A T I C S$

Identify Repeating Letters

Identify Unique Letters

Defining Selection Strategy

Case 1: All 5 Distinct Letters

Calculate Case 1

Case 2: 1 Pair and 3 Distinct Letters

Calculate Case 2

Case 3: 2 Pairs and 1 Distinct Letter

Calculate Case 3

Final Summation

The Sigma Insight: Combinations and Selection

Analyzing the Anatomy of MATHEMATICS

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

The Grand Finale

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