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91Ó°ÊÓ

Consider lists made from the letters \(T, H, E, O, R, Y,\) with repetition allowed. (a) How many length-4 lists are there? (b) How many length-4 lists are there that begin with \(T ?\) (c) How many length- 4 lists are there that do not begin with \(T\) ?

Short Answer

Expert verified
The total number of length-4 lists is 1296. The number of lists beginning with T is 216. The number of lists not beginning with T is 1080.

Step by step solution

01

Calculate no restriction lists

Using the formula for permutations with repetition, which is \(n^r\), where \(n\) is the number of items to choose from (in this case, 6 letters), and \(r\) is how many we choose (4 in this case), the total number of possible 4-letter combinations is \(6^4 = 1296\).
02

Calculate list that begin with T

In this case, the first letter is always T, so there is only one choice for the first letter. However, for the remaining 3 letters, we still have 6 choices each (as repetition is allowed). So, using the formula mentioned above, it's \(1*6^3 = 216\).
03

Calculate lists that do not begin with T

In this case, we cannot use T as the first letter, so we only have 5 options for the first letter. However, for the remaining 3 letters, we still have 6 choices each. So, using the formula mentioned above, it's \(5*6^3 = 1080\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Permutations
Permutations are all about arranging items in a specific order. When we talk about permutations, we're interested in how many different ways we can order a set of items. Consider a word like "THEORY". It contains six different letters, and if we were to arrange all of them, any change in their order creates a new permutation. Permutations are crucial when the order matters. For example, each time we rearrange the letters of "THEORY," we get a different permutation. This concept is deeply connected to the idea that each unique order or sequence counts as a separate arrangement.

When repetitions are allowed, as in the exercise, the complexity increases. We aren't just rearranging unique items; we're selecting them repeatedly. This is where the formula for permutations with repetition, such as \(n^r\), becomes very handy. The formula \(n^r\) computes the number of possible sequences where \(n\) is the number of items to choose from, and \(r\) is the selected length. In our example, this results in \(6^4 = 1296\), representing all possible 4-letter lists from "THEORY", with repetition allowed.
Combinatorics
Combinatorics is the study of counting, arranging, and finding patterns. It helps us analyze scenarios to see how combinations of items form. However, unlike permutations, in combinatorics, order does not necessarily matter unless specified. The focus shifts from ordering to selecting, and sometimes you may have conditions, like allowing repetitions or excluding specific items.

In our exercise, combinatorics helps us determine how many different ways we can form lists from a set of letters. For instance, part (a) asked us how many lists can be made without any restrictions, whereas part (c) brought in a restriction—starting the list without a specific letter, 'T'. This kind of problem-solving is at the heart of combinatorics, where understanding how to approach and solve different selection scenarios without and with conditions is key.
Lists with Repetition
Lists with repetition are collections where items can repeat. In scenarios like the exercise example, repetition greatly expands possible combinations. Imagine you have a bowl with different types of candy, and you're allowed to take the same candy more than once. This situation is similar to our exercise, where lists can reuse the same letter multiple times.

The formula \(n^r\), as mentioned before, is what we use when repetition is allowed. It's important because it changes the dynamics of how combinations are calculated. For example, for a 4-letter list from "THEORY," repetition means that each spot in the list can be any of the six letters. The first spot could have a 'T', the next could also have a 'T', and so on, for every possible position in the list. This is a key difference from situations where items cannot repeat, as it multiplies the possibilities extensively. This characteristic makes problems involving lists with repetition unique and often more challenging.

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Most popular questions from this chapter

You deal a pile of cards, face down, from a standard 52 -card deck. What is the least number of cards the pile must have before you can be assured that it contains at least five cards of the same suit?

Three people in a group of ten line up at a ticket counter to buy tickets. How many lineups are possible?

This problem concerns lists made from the letters \(A, B, C, D, E, F, G, H, I, J .\) (a) How many length- 5 lists can be made from these letters if repetition is not allowed and the list must begin with a vowel? (b) How many length- 5 lists can be made from these letters if repetition is not allowed and the list must begin and end with a vowel? (c) How many length- 5 lists can be made from these letters if repetition is not allowed and the list must contain exactly one \(A ?\)

Consider 4-card hands dealt off of a standard 52-card deck. How many hands are there for which all 4 cards are of the same suit or all 4 cards are red? Solution: Let \(A\) be the set of 4 -card hands for which all four cards are of the same suit. Let \(B\) be the set of 4 -card hands for which all four cards are red. Then \(A \cap B\) is the set of 4 -card hands for which the four cards are either all hearts or all diamonds. The answer to our question is \(|A \cup B|=|A|+|B|-|A \cap B|=\) \(4\left(\begin{array}{c}13 \\\ 4\end{array}\right)+\left(\begin{array}{c}26 \\\ 4\end{array}\right)-2\left(\begin{array}{c}13 \\\ 4\end{array}\right)=2\left(\begin{array}{c}13 \\\ 4\end{array}\right)+\left(\begin{array}{c}26 \\\ 4\end{array}\right)=1430+14,950=\mathbf{1 6}, \mathbf{3 8 0}\).

You roll a dice six times in a row. How many possible outcomes are there that have two 1's three 5 's and one 6 ?

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