Problem 58 - Algebra for College Students

Question

Each arrangement of the 11 letters of the word MISSISSIPPI is put on a slip of paper and placed in a hat. One slip is drawn at random from the hat. Find the probability that the slip contains an arrangement of the letters with the four S's at the beginning. \(\frac{1}{330}\)

Step-by-Step Solution

Answer

The probability is \(\frac{1}{330}\).

Step 1: Identify Total Arrangements

First, determine the total possible arrangements of the letters in the word 'MISSISSIPPI.' The word has 11 letters, with repetitions of certain letters: 1 M, 4 I's, 4 S's, and 2 P's. Use the formula for permutations of multiset: \[ \frac{11!}{1!4!4!2!} \].

Step 2: Calculate Total Arrangements

Calculate the total arrangements by substituting into the formula: \( \frac{11!}{1!4!4!2!} = \frac{39916800}{1 \times 24 \times 24 \times 2} = 34650 \).

Step 3: Identify Favorable Arrangements

Next, identify the arrangements with the four S's at the beginning. Treat the remaining 7 letters (MIIPPI) as needing arrangement. Find the permutations of these 7 letters: \[ \frac{7!}{1!4!2!} \].

Step 4: Calculate Favorable Arrangements

Calculate the arrangements with the S's at the beginning: \( \frac{7!}{1!4!2!} = \frac{5040}{1 \times 24 \times 2} = 105 \).

Step 5: Find Probability

Finally, the probability of a slip having the S's at the beginning is the ratio of favorable to total arrangements: \( \frac{105}{34650} = \frac{1}{330} \).

Key Concepts

Permutation of Multiset

When dealing with permutations of multisets, we are arranging a set of items where some items are identical. Here, the key is to recognize the repetitions and adjust the typical permutation formula accordingly. The formula for the permutations of a multiset of size \( n \) with elements repeated \( a, b, c, \ldots \) times respectively is:

• \[ \frac{n!}{a!b!c!\ldots} \]

Given the exercise, we have the word 'MISSISSIPPI' which consists of 11 letters including repeated letters: 1 M, 4 I's, 4 S's, and 2 P's.
Hence, the formula becomes \( \frac{11!}{1!4!4!2!} \). This approach helps in calculating the number of unique arrangements such as for multiset letters, where identical letters are incorrectly counted as unique if not factored in.
Such permutations are crucial in solving many probability problems where items repeat.

Probability Calculation

Probability in permutations often involves finding the number of successful or favorable outcomes divided by the total number of possible outcomes. This is evident in the problem where we aim to find the arrangement probability with four S's at the start.We first calculate all possible permutations without any additional restriction, which we did using:

• Total arrangements: \( \frac{11!}{1!4!4!2!} = 34650 \)

Next, find how many permutations match our specific criterion (4 S's at the beginning) and calculate:

• Favorable arrangements: \( \frac{7!}{1!4!2!} = 105 \)

Thus, the probability of the letters being in an arrangement with 4 S's at the beginning is \( \frac{105}{34650} = \frac{1}{330} \).
This ratio effectively demonstrates the proportion of favorable outcomes in the entire set of possibilities.

Arrangement of Letters

An arrangement of letters in a word like 'MISSISSIPPI' with repeated letters, requires careful counting to avoid overcounting arrangements that look identical due to repetition.❖ **Understanding the Process**:

• Identify total and type of letters (e.g., 4 I's, 4 S's).
• Calculate total permutations: \( \frac{11!}{1!4!4!2!} \).

One should remember that even with all letters being arranged differently, indistinguishable letters (same letters) affect the count.
This explains why using the permutation of multiset is necessary.
❖ **Special Arrangements**:

• Consider special cases, such as having a certain sequence. Here 4 S's are at the start.
• Treat the rest of the letters (7 remaining) as a separate permutation case: \( \frac{7!}{1!4!2!} \).

Arranging letters with specific positions concatenates combinatorial thinking with creative problem solving, essential in probability and combinatorics.