If the number of words, with or without meaning, which can be made using all the letters of the word

Q: Wait, why do we use the Gap Method instead of just subtracting the 'together' cases from the total?

Both methods are mathematically valid! However, the Gap Method is often less prone to errors in complex problems. Subtracting 'together' cases requires you to treat 'CS' as a single unit, which can get tricky with identical letters. The Gap Method is a direct, elegant way to ensure separation without needing to calculate the total set first.

Q: Why do we divide by $2!$ three times in the base arrangement?

Great catch! In the word 'MATHEMATICS', the letters M, A, and T each appear twice. When we arrange these 9 base letters, we must account for the fact that swapping two identical 'M's, 'A's, or 'T's does not create a new, unique word. Dividing by $2!$ for each pair removes these redundant, identical arrangements.

Q: Why is it ${}^{10}P_2$ and not ${}^{10}C_2$?

This is a classic trap! We use ${}^{10}P_2$ because the order of 'C' and 'S' matters. Placing 'C' in the first gap and 'S' in the second gap creates a different word than placing 'S' in the first gap and 'C' in the second. Since 'C' and 'S' are distinct entities, we must account for their arrangement, hence permutations ($P$) rather than combinations ($C$).

Question

If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which $C$ and $S$ do not come together, is $(6!)k$ then $k$ is equal to

Options:

(A) 2835

(B) 5670

(C) 1890

(D) 945

Sigma Percentile · Accepted Answer

### Analyzing the Architecture of Permutations

Welcome, fellow explorer of the mathematical universe! Today, we are going to dissect a classic JEE Advanced combinatorics problem. We are looking at the word 'MATHEMATICS' and imposing a strict social distancing rule: the letters 'C' and 'S' must never sit together.

It sounds like a simple constraint, but it opens the door to a beautiful, structured way of thinking about arrangements.

### Phase 1: The Frequency Breakdown

Before we dive into the logic, we must understand our building blocks. The word 'MATHEMATICS' has $11$ letters.

If we were to just scramble them, we would have a massive number of possibilities. However, we have identical letters, which act as 'twins' in our permutation.

Let's tally them up: we have two M's, two A's, and two T's. The remaining letters—H, E, I, C, and S—are unique. This frequency breakdown is our foundation, and it reminds us that in the world of combinatorics, we must always account for indistinguishable items to avoid overcounting.

### Phase 2: The Gap Method Strategy

Now, how do we enforce the rule that 'C' and 'S' must not be together? We use the 'Gap Method'.

Imagine the other $9$ letters standing in a line. They create 'gaps'—one at the very beginning, one between each letter, and one at the very end.

If we place 'C' and 'S' into these gaps, they will always be separated by at least one other letter. It is a foolproof strategy that turns a complex constraint into a simple placement problem.

### Phase 3: Arranging the Base

First, let's arrange our $9$ base letters (M, M, A, A, T, T, H, E, I). The number of unique ways to arrange these is given by the formula for permutations of a multiset:

$$
\frac{9!}{2! 	imes 2! 	imes 2!}
$$

Don't let the factorial notation intimidate you; it is just a way of saying 'all possible arrangements divided by the ways to swap the identical letters'.

### Phase 4: The Art of Gap Placement

With our $9$ base letters in a row, how many gaps do we have? As we visualized, $9$ letters create $10$ gaps.

We need to place our two distinct letters, 'C' and 'S', into these $10$ gaps. Since 'C' and 'S' are distinct, the order in which we place them matters.

This is a permutation problem: we need to choose $2$ gaps out of $10$ and arrange 'C' and 'S' within them. This is calculated as:

$$
{}^{10}P_2 = 10 	imes 9 = 90
$$

We have $90$ ways to place our 'C' and 'S' safely.

### Phase 5: The Final Synthesis

Now, we combine our two independent choices. The total number of valid words is the product of the base arrangements and the gap placements:

$$
	ext{Total Ways} = \frac{9!}{2! 	imes 2! 	imes 2!} 	imes 90
$$

We know that $2! 	imes 2! 	imes 2! = 8$. So, our expression becomes $\frac{9!}{8} 	imes 90$.

The problem asks us to express this as $(6!)k$. Let's expand $9!$ partially to reveal the $6!$: $9! = 9 	imes 8 	imes 7 	imes 6!$. Substituting this back, we get:

$$
\frac{9 	imes 8 	imes 7 	imes 6!}{8} 	imes 90
$$

The $8$ in the numerator and denominator cancels out beautifully, leaving us with:

$$
(9 	imes 7 	imes 90) 	imes 6!
$$

Calculating $9 	imes 7 	imes 90$, we get $63 	imes 90 = 5670$. Thus, our total number of words is $5670 	imes 6!$.

Comparing this to $(6!)k$, we find that **$k = 5670$**. We have navigated the complexity and arrived at the elegant solution.

Animated Solution for Mathematics - Permutations and Combinations: If the number of words, with or without meaning, which can be made using all the letters of the word MATHEMATICS in which $C$ and $S$ do not come together, is $(6!) k$ then $k$ is equal to

Select Answer:

Visualized Solution

Analyzing the Word

The Constraint & Strategy

Arranging the Base Letters

Identifying the Gaps

Placing $C$ and $S$

Combining the Arrangements

Simplifying the Expression

Final Calculation of $k$

The Sigma Insight: Linear Permutations

Analyzing the Architecture of Permutations

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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