Problem 54 - A Complete Resource Book in Mathematics for JEE Main

Question

How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two \(S\) are adjacent? (A) \(8 \cdot{ }^{6} C_{4} \cdot{ }^{7} C_{4}\) (B) \(6 \cdot 7 \cdot{ }^{8} C_{4}\) (C) \(6 \cdot 8 \cdot{ }^{7} C_{4}\) (D) \(7 \cdot{ }^{6} C_{4} \cdot{ }^{8} C_{4}\)

Step-by-Step Solution

Answer

The number of different words is \(6 \cdot 7 \cdot {}^{8}C_4\).

Step 1: Count Total Letters in 'MISSISSIPPI'

'MISSISSIPPI' consists of 11 letters, predominantly 'I', 'S', and 'P'. Specifically: - 'M' appears 1 time, - 'I' appears 4 times, - 'S' appears 4 times, - 'P' appears 2 times.

Step 2: Arrange Letters 'M', 'I', 'P' Without 'S'

Let's focus on arranging the non-'S' letters: 'M', 'I', 'I', 'I', 'I', 'P', 'P'.- Count of letters = 7- Total arrangements = \(\frac{7!}{1!4!2!}\)- Calculation: \(\frac{5040}{48} = 105\).So, we have 105 arrangements without considering 'S's.

Step 3: Insert 'S's Between and Around Arranged Letters

Four 'S's must be placed around and between these non-'S' letters:- Potential slots: _ M _ I _ I _ I _ I _ P _ P _- Total slots available = 8 (includes beginning and end slots)- Select positions for 4 'S's from these 8 slots: \({}_{8}C_{4}\).- Calculation: \(\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70\).

Step 4: Calculate Total Arrangements

Combine the arrangements from Step 2 and Step 3 using multiplication:- Total different words = 105 (arrangements of non-'S' letters) \(\times\) 70 (ways to place 'S's)- Calculation: 105 \(\times\) 70 = 7350.

Key Concepts

Permutation and Combination

Permutation and combination are both crucial concepts in combinatorics, serving as methods to count different ways of arranging or selecting items. These concepts help us determine the order and selection in various situations.

**Permutation** refers to the arrangement of items in a particular sequence or order. When dealing with permutations, order matters. For instance, in the word MISSISSIPPI, arranging the letters involves permutations of these letters. The computation for arranging letters takes into account repeated characters by dividing by their respective factorials to avoid overcounting identical permutations.
* Example calculation: Suppose we arrange the letters 'M', 'I', 'I', 'I', 'I', 'P', 'P'. The permutations are calculated as: - \(\frac{7!}{1!4!2!}\), taking into account the identical 'I's and 'P's.

**Combination**, on the other hand, involves selecting items where order does not matter. It is often used in problems where groups or teams are formed. In this exercise, to determine the placement of 'S's without them being adjacent, combinations are used to select slots among available positions.

Factorial Calculation

The factorial is a fundamental operation in combinatorics. It is the product of all positive integers up to a given number, denoted as \( n! \). Factorials are used to compute permutations, combinations, and arrangements in various problems.

For example, \(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\).
Factorials help in determining the total number of permutations for an arrangement of distinct items. When some items are repeated, as with the 'I's and 'P's in our word, their factorials divide the total to avoid overcounting equivalent sequences.

In the solution process, once you calculate the factorial of a total number of items, you adjust for any repetitions by dividing through by the factorial of the number of repeated items: \[ \frac{7!}{1!4!2!} = 105 \].
This calculation is essential to determine the right number of distinct permutations among repeated characters in a sequence.

Non-Adjacent Arrangement

A non-adjacent arrangement involves positioning items in such a way that no two specified items are next to each other. This concept commonly appears in problems where restrictions are applied to the placement of certain elements.

In our example involving the word MISSISSIPPI, 'S's need to be arranged such that no two 'S's are adjacent. This requires a thoughtful placement approach to ensure sufficient gaps between 'S's.
To achieve this, we first arrange all letters except 'S's and calculate the different placements for the remaining letters. For steps like this:

• We arrange non-'S' letters first, creating gaps for possible 'S' placement.
• We then count available gaps or slots, including positions before and after the sequence. This creates spaces where 'S's can fit without being next to each other.
• We use combinations to choose positions for 'S's from these gaps, ensuring they are spread without any two being adjacent, calculated as \(^{8}C_{4}\).

Such arrangements emphasize the careful positioning needed to satisfy conditions of separation and highlight the utility of combination methods where order is not a priority but position selection is crucial.