Problem 23
Question
Each arrangement of the six letters of the word CYCLIC is put on a slip of paper and placed in a hat. One slip is drawn at random. Find the probability that the slip contains an arrangement with the \(Y\) at the beginning. \(\frac{1}{6}\)
Step-by-Step Solution
Verified Answer
The probability is \( \frac{1}{6} \).
1Step 1: Determine Total Arrangements
The word CYCLIC consists of six letters. The total number of arrangements, considering the repeated "C", can be calculated using the formula for permutations of multiset: \( \frac{n!}{p!} \), where \( n \) is the total number of letters and \( p \) is the factorial of the number of times any letter is repeated. Here, n = 6 and the letter C is repeated twice. Thus, total arrangements = \( \frac{6!}{2!} = \frac{720}{2} = 360 \).
2Step 2: Find Arrangements with Y at the Beginning
If Y is at the beginning, we have the remaining letters: C, C, L, I, and C. We now calculate the permutations of these 5 letters. Since C is still repeated twice, the number of arrangements is \( \frac{5!}{2!} = \frac{120}{2} = 60 \).
3Step 3: Calculate the Probability
The probability that a randomly drawn slip has Y at the beginning is the number of favorable outcomes (arrangements with Y at the start) divided by the total number of arrangements. Thus, the probability is \( \frac{60}{360} = \frac{1}{6} \).
Key Concepts
PermutationsFactorialArrangement of Letters
Permutations
Permutations are all about arranging things in different ways. They help us figure out how many ways we can arrange a set of items. Think of it as trying to figure out the different ways to line up books on a shelf. If you have a group of items, and you want to find out all possible ways to arrange them, you're dealing with a permutation.
In permutations, the order in which things are arranged matters. For example, the arrangement "ABC" is different from "CBA". The formula for permutations when every item is unique is simply the factorial of the number of items, written as \[ n! \]. If, however, there are repeating items, like letters in a word, we adjust this by dividing by the factorial of the number of repetitions. For instance, in the word "CYCLIC", the letter "C" appears twice. Therefore, we use the formula: \[ \frac{n!}{p!} \], where \( n! \) is the factorial of the total number of items, and \( p! \) is the factorial of the number of repeating items.
In permutations, the order in which things are arranged matters. For example, the arrangement "ABC" is different from "CBA". The formula for permutations when every item is unique is simply the factorial of the number of items, written as \[ n! \]. If, however, there are repeating items, like letters in a word, we adjust this by dividing by the factorial of the number of repetitions. For instance, in the word "CYCLIC", the letter "C" appears twice. Therefore, we use the formula: \[ \frac{n!}{p!} \], where \( n! \) is the factorial of the total number of items, and \( p! \) is the factorial of the number of repeating items.
Factorial
A factorial is a special function in mathematics represented by an exclamation mark (\(!\)) next to a number. It essentially means multiplying that number by every whole number less than itself down to one. So \( n! = n \times (n-1) \times (n-2) \times ... \times 1 \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
Factorials are incredibly useful in permutations and combinations because they calculate the total number of possible ways to arrange a set of items. They play a key role when order matters. In problems like arranging letters or finding possible outcomes in games or puzzles, factorials make calculations simpler.
Remember, factorial of 0 is 1, (\(0! = 1\)), which is a special rule applied in mathematical formulae.
Factorials are incredibly useful in permutations and combinations because they calculate the total number of possible ways to arrange a set of items. They play a key role when order matters. In problems like arranging letters or finding possible outcomes in games or puzzles, factorials make calculations simpler.
Remember, factorial of 0 is 1, (\(0! = 1\)), which is a special rule applied in mathematical formulae.
Arrangement of Letters
The arrangement of letters in a word considers the different ways to organize the letters, especially noting if some are repeated. When arranging letters like in "CYCLIC", it's more complex because the letter "C" repeats. To find the number of distinct arrangements, we use the permutation formula for multisets, \[ \frac{n!}{p!} \], where \( n \) is the total number of letters and \( p \) is the number of times a letter repeats.
When asked to find arrangements with specific conditions, like "Y" at the beginning, we only focus on arranging the other letters. Here, "Y" is fixed at the start and the rest are arranged, taking account of any repeated letters. By calculating these specific arrangements and comparing them to the total arrangements, we can determine probabilities. So knowing how to handle letter arrangements efficiently requires considering both total possibilities and any constraints or repeats that may exist.
When asked to find arrangements with specific conditions, like "Y" at the beginning, we only focus on arranging the other letters. Here, "Y" is fixed at the start and the rest are arranged, taking account of any repeated letters. By calculating these specific arrangements and comparing them to the total arrangements, we can determine probabilities. So knowing how to handle letter arrangements efficiently requires considering both total possibilities and any constraints or repeats that may exist.
Other exercises in this chapter
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