How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no t

Q: Why can't we just use the total permutations of MISSISSIPPI and subtract the cases where S's are together?

That is a common trap! While subtraction works for two items, it becomes incredibly complex for four items. You would need to use the Principle of Inclusion-Exclusion to account for cases where two S's are together, three are together, or all four are together in various configurations. The Gap Method is much cleaner and avoids this headache entirely.

Q: Why do we use ${}^8C_4$ and not ${}^8P_4$ for placing the S's?

Great question. We use combinations (${}^8C_4$) because the S's are identical. If you swap the position of two S's, the word remains exactly the same. Therefore, the order in which you place the S's into the chosen gaps does not matter, only which gaps you choose.

Q: How do we know there are exactly 8 gaps?

Think of the 7 non-S letters as barriers: _ L _ L _ L _ L _ L _ L _ L _. If you count the underscores, you will see one before the first letter, one after the last letter, and one between each of the 7 letters. That is $7 + 1 = 8$ total slots available.

Question

How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?

Options:

(A) 8 \cdot ^6C_4 \cdot ^7C_4

(B) 6 \cdot 7 \cdot ^8C_4

(C) 6 \cdot 8 \cdot ^7C_4

(D) 7 \cdot ^6C_4 \cdot ^8C_4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The word **MISSISSIPPI** consists of 11 letters in total: one $M$, four $I$'s, two $P$'s, and four $S$'s. The objective is to arrange these letters such that no two $S$'s are adjacent.

This is a classic combinatorial problem that is best solved using the **Gap Method**. By isolating the constrained elements, we transform a complex arrangement problem into a structured, logical sequence.

### The Strategy: Divide and Conquer

To apply the Gap Method, we first ignore the four $S$'s and focus on the remaining letters: $M$, $I$, $I$, $I$, $I$, $P$, and $P$. There are 7 such letters in total.

We arrange these 7 letters using the formula for permutations of a multiset. The number of ways to arrange these letters is given by:

$$ \frac{7!}{4!2!1!} = \frac{7!}{4!2!} $$

### The Elegant Transformation

To simplify the expression for the arrangement of non-$S$ letters, we perform the following calculation:

$$ \frac{7 	imes 6!}{4!2!} = 7 	imes \frac{6!}{4!2!} $$

Recognizing that $\frac{6!}{4!2!}$ is equivalent to the combination formula ${}^6C_4$, we simplify our arrangement count to $7 	imes {}^6C_4$. Keeping an eye on the structure of the expression is essential for matching potential problem options.

### Creating the Gaps

With the 7 non-$S$ letters arranged, we create gaps between them to place the $S$'s. Representing the letters as $L$, the arrangement looks like this:

$$ \_ L \_ L \_ L \_ L \_ L \_ L \_ L \_ $$

Because there are 7 letters, there are 8 available gaps (one at each end and six between the letters). To ensure no two $S$'s are adjacent, we must place each of our 4 identical $S$'s into a distinct gap.

The number of ways to choose 4 gaps out of the 8 available is given by the combination:

$$ {}^8C_4 $$

### The Final Synthesis

By the **Fundamental Principle of Counting**, we multiply the number of ways to arrange the non-$S$ letters by the number of ways to choose the gaps for the $S$'s.

The total number of valid arrangements is:

$$ (7 	imes {}^6C_4) 	imes {}^8C_4 $$

This result represents the complete solution to the problem. By isolating the constrained elements, we have successfully navigated the complexity of the word **MISSISSIPPI**.

Animated Solution for Mathematics - Permutations and Combinations: How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?

Select Answer:

Visualized Solution

Analyze the Word $M I S S I S S I P P I$

The Constraint

Arrange Non- $S$ Letters

Permutations of Non- $S$ Letters

Match with Options Format

Create Gaps

Place the $S$ Letters

Gap Combinations

Final Answer

The Sigma Insight: Linear Permutations

Analyzing the Setup

The Strategy

The Elegant Transformation

Creating the Gaps

The Final Synthesis

Similar Questions

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

The number of words (with or without meaning) that can be formed from all the letters of the word "LETTER" in which vowels never come together is

The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is

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

The number of six letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is .

The number of arrangements of the letters of the word "INDEPENDENCE" in which all the vowels always occur together is

The number of words, with or without meaning, that can be formed using all the letters of the word ASSASSINATION so that the vowels occur together, is ____.

The number of arrangements of the letters of the word BANANA in which the two N's do not appear adjacently is

How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order

The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is :