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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 one another (such as in \(B B B A A A D D D C C C\) ?

Step by step solution

01

Count Total Arrangements with Distinguishable Molecules

Assume each molecule type (A, B, C, D) is distinguishable (e.g., A1, A2, A3, etc.). The total number of ways to arrange the 12 molecules (3 of each type) is given by a permutation formula: \[\frac{12!}{3! \times 3! \times 3! \times 3!}\] Calculate this to determine the baseline number of arrangements when molecules of the same type are distinguishable.

02

Calculate the Total Arrangements without Distinguishable Molecules

Now, consider the molecules of the same type (e.g., A's) to be identical. Since the molecules of each type are not distinguishable, the total number of arrangements does not change from the distinguishable case but remains as in Step 1. Since 3! ways are accounted for in each type already, this initial calculation suffices.

03

Probability of Adjacent Molecules of Each Type

To find the probability that all three molecules of each type are adjacent, first treat each triplet as a single "block". Thus, four blocks need arranging: \[ \{AAA, BBB, CCC, DDD\} \]. There are 4! ways to arrange these blocks.The probability is the ratio of favorable outcomes to total outcomes calculated in Step 2:\[ \frac{4!}{\frac{12!}{3! \times 3! \times 3! \times 3!}} \] Evaluate this probability.

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Key Concepts

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

Permutations

In combinatorics, permutations refer to the different ways in which a set of objects can be arranged. When calculating permutations, the order of arrangement is crucial. For the problem at hand, consider that we have 12 molecules to arrange, with three of each type (A, B, C, and D). We start by assigning unique labels to each molecule, treating them as distinct. This transforms the problem into arranging 12 unique items. The formula for permutations where some items are identical is:

• The general formula used is: \[\frac{n!}{n_1! \times n_2! \times n_3! \times n_4!}\].Where \(n\) is the total number of items and \(n_1, n_2, \) etc., represent the identical items in each subset.

Applying this, we get the total number of permutations as \(\frac{12!}{3! \times 3! \times 3! \times 3!}\). This result gives us how many ways the 12 molecules can be arranged when considering indistinguishable cases separated by type.

Probability

Probability in this context measures the likelihood of a specific arrangement happening among many possibilities. For our problem, we want to calculate the probability that all molecules of the same type end up adjacent. Each group of three molecules is treated as a single block making them four blocks to arrange. The favorable arrangements (with molecules clustered together) are found by arranging these blocks, which is calculated as:

• There are 4! ways to arrange these blocks since they are considered as units or "super-molecules".

The ratio of these favorable arrangements to the total number of arrangements gives the probability. Thus, we calculate this by:\[\text{Probability} = \frac{4!}{\frac{12!}{3! \times 3! \times 3! \times 3!}}\].This formula evaluates the chances of having triplets of the same type in sequence.

Arrangements

Arrangements refer to the different ways molecules can be positioned in a sequence. Here we consider the molecules to be linked in a chain, which changes the complexity of arrangements since they are not just placed separately but connected in a series. The concept of arrangements in this context includes recognizing that molecules originally seen as identical can be considered as distinguishable when labeling them in calculations.

• Using the factorial approach \(\frac{12!}{3! \times 3! \times 3! \times 3!}\), the task accounts for similarity by dividing out the identical ways to arrange the same type of molecules.

The problem of arranging 12 molecules with repeated items is a classic case of arranging sequences with constraints, commonly solved using permutations.

Distinct objects

In the context of this exercise, the molecules initially considered as distinct become indistinguishable under certain conditions, such as lacking specific labels. Understanding distinct versus indistinct involves knowing when to label entities (like molecules) the same or differently based on situational changes.

• If every molecule was seen as distinct, each one would have a unique identity, providing more permutation possibilities.
• However, our exercise asks us to treat the same type of molecules as indistinct after accounting for types initially treated as distinct.

This distinction affects both permutations and probability since initial computation assumes distinction, while results depend on treating like molecules as indistinguishable upon final calculation.