/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Three molecules of type \(A\), t... [FREE SOLUTION] | 91Ó°ÊÓ

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Three molecules of type \(A\), three of type \(B\), three of type \(C\), and three of type \(D\) are to be linked together to form a chain molecule. One such chain molecule is \(A B C D A B C D A B C D\), and another is \(B C D D A A A B D B C C\). a. How many such chain molecules are there? [Hint: If the three A's were distinguishable from one another- \(A_{1}, A_{2}, A_{3}\)-and the \(B\) 's, \(C\) 's, and \(D\) 's were also, how many molecules would there be? How is this number reduced when the subscripts are removed from the \(A\) 's?] b. Suppose a chain molecule of the type described is randomly selected. What is the probability that all three molecules of each type end up next to each other (such as in \(B B B A A A D D D C C C\) )?

Short Answer

Expert verified
a. Total distinct molecules: 369600. b. Probability of same type together: \(\frac{1}{3696}\).

Step by step solution

01

Count Total Arrangements with Distinguishable Molecules

First, we treat each molecule as unique by adding subscripts: - A's: \(A_1, A_2, A_3\) - B's: \(B_1, B_2, B_3\) - C's: \(C_1, C_2, C_3\) - D's: \(D_1, D_2, D_3\)This gives us 12 unique positions for these molecules. The total arrangements can be calculated using permutations of the 12 unique items: \[ 12! \]
02

Adjust for Indistinguishability

Now, we account for the fact that the molecules of type A, B, C, and D are indistinguishable among themselves. This means we must divide by the number of ways to arrange each group of identical molecules:- 3! arrangements for the A's- 3! arrangements for the B's- 3! arrangements for the C's- 3! arrangements for the D'sSo, the total number of distinct chain molecules is: \[ \frac{12!}{3! \times 3! \times 3! \times 3!} \]
03

Calculate Specific Arrangements with Molecules Adjacent

To find the number of arrangements where all molecules of the same type are adjacent (e.g., BBB AA DDD CCC), treat each group of three as a single 'block'. This creates 4 blocks, which can be arranged in \(4!\) different ways. Additionally, the 3 molecules within each block can be arranged within themselves in \(3!\) ways. Therefore:The number of arrangements where all the same molecules are together is:\[ 4! \times (3!)^4 = 24 \times 6^4 \]
04

Calculate the Probability

The probability that all the molecules of each type end up next to each other is obtained by dividing the number of such arrangements by the total arrangements calculated in Step 2:\[ \frac{4! \times (3!)^4}{\frac{12!}{3! \cdot 3! \cdot 3! \cdot 3!}} \]

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Key Concepts

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

Permutation
In combinatorics, a permutation refers to the arrangement of objects in a specific order. This concept becomes handy when determining the number of possible configurations of a set of unique items. For example, arranging 12 distinguishable molecules (like labeled molecules in the exercise) requires calculating the permutation of these 12 items, which is computed as 12 factorial, denoted as \( 12! \).
When considering permutation, it's crucial to remember that each different order of these molecules represents a completely unique permutation. In simpler terms, swapping the position of any two molecules results in a new permutation.
Understanding permutations is essential because, in real-life scenarios, molecules such as those in the problem might be arranged differently based on conditions like temperature or pressure, leading to a different structure with unique properties. Permutation provides a foundation for calculating such arrangements in a systematic manner.
Probability
Probability is a measure of the likelihood that a particular event will occur. In the context of the exercise, it answers, "What are the chances that molecules of the same type will all be next to each other when randomly linked?".
To find this probability, divide the number of favorable arrangements (where molecules group together) by the total number of possible arrangements. This ratio gives a value between 0 and 1, indicating the probability.
  • A probability closer to 1 signifies a higher chance of the event occurring.
  • Conversely, a probability near 0 suggests it's unlikely to happen.

In our exercise, the probability that all three molecules of each type are adjacent involves comparing the specific grouped arrangements to the total permutations without group restrictions.
Indistinguishability
Indistinguishability in combinatorics deals with cases where objects look identical and switching them doesn't create a new arrangement. This concept comes into play when calculating the total number of unique permutations.
For instance, even though there are 12 positions for molecules, simply permuting them under the impression that all are unique (using \( 12! \)) isn’t accurate. Each group of three identical molecules, like the three A's, can be rearranged among themselves without generating a new configuration. Thus, we divide \( 12! \) by \( 3! \times 3! \times 3! \times 3! \), acknowledging that these similar molecules' arrangements don't count as new permutations.
Understanding indistinguishability helps in solving real-world problems like determining the number of unique necklaces that can be made from identical beads.
Factorial
The factorial of a number, symbolized by an exclamation point (!), is a product of all positive integers up to that number. For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). It's a widely-used mathematical concept in permutations and combinations.
In the exercise, the factorial is used multiple times:
  • \( 12! \) represents the permutation of all 12 distinguishable molecules.
  • \( 3! \) accounts for the arrangements within each subgroup of identical molecules.

Factorials play a critical role in combinatorial calculations, especially those dealing with complex arrangements of items, as it's a concise way of calculating permutations and accommodating indistinguishable elements in a set.

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Most popular questions from this chapter

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \(C_{1}=\\{\) left ear tag is lost \(\\}\) and \(C_{2}=\\{\) right ear tag is lost . Let \(\pi=P\left(C_{1}\right)=P\left(C_{2}\right)\), and assume \(C_{1}\) and \(C_{2}\) are independent events. Derive an expression (involving \(\pi\) ) for the probability that exactly one tag is lost given that at most one is lost ("Ear Tag Loss in Red Foxes," J. Wildlife Manag., 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

Suppose a single gene determines whether the coloring of a certain animal is dark or light. The coloring will be dark if the genotype is either \(A A\) or \(A a\) and will be light only if the genotype is \(a a\) (so \(A\) is dominant and \(a\) is recessive). Consider two parents with genotypes \(A a\) and \(A A\). The first contributes \(A\) to an offspring with probability \(1 / 2\) and \(a\) with probability \(1 / 2\), whereas the second contributes \(A\) for sure. The resulting offspring will be either \(A A\) or \(A a\), and therefore will be dark colored. Assume that this child then mates with an \(A a\) animal to produce a grandchild with dark coloring. In light of this information, what is the probability that the first-generation offspring has the \(A a\) genotype (is heterozygous)? [Hint: Construct an appropriate tree diagram.]

Consider four independent events \(A_{1}, A_{2}, A_{3}\), and \(A_{4}\) and let \(p_{i}=P\left(A_{i}\right)\) for \(i=1,2,3,4\). Express the probability that at least one of these four events occurs in terms of the \(p_{i}\) 's, and do the same for the probability that at least two of the events occur.

An aircraft seam requires 25 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with the same probability. a. If \(20 \%\) of all seams need reworking, what is the probability that a rivet is defective? b. How small should the probability of a defective rivet be to ensure that only \(10 \%\) of all seams need reworking?

The College of Science Council has one student representative from each of the five science departments (biology, chemistry, statistics, mathematics, physics). In how many ways can a. Both a council president and a vice president be selected? b. A president, a vice president, and a secretary be selected? c. Two members be selected for the Dean's Council?

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