Problem 43 - Applied Statistics and Probability for Engineers

Question

In the laboratory analysis of samples from a chemical process, five samples from the process are analyzed daily. In addition, a control sample is analyzed twice each day to check the calibration of the laboratory instruments. (a) How many different sequences of process and control samples are possible each day? Assume that the five process samples are considered identical and that the two control samples are considered identical. (b) How many different sequences of process and control samples are possible if we consider the five process samples to be different and the two control samples to be identical? (c) For the same situation as part (b), how many sequences are possible if the first test of each day must be a control sample?

Step-by-Step Solution

Answer

(a) 21 sequences, (b) 2520 sequences, (c) 120 sequences.

Step 1: Understand the Concept of Identical Samples

When samples are identical, it means that swapping two samples doesn't result in a new sequence because the samples are indistinguishable from each other. This affects the sequence calculations because permutations of identical items will not count as distinct sequences.

Step 2: Calculate the Number of Sequences for Identical Samples

For part (a), we have an overall sequence containing 5 identical process samples and 2 identical control samples. The total sequence has 7 positions. The formula to find the number of sequences is given by \( \frac{7!}{5! \times 2!} \). This formula divides the total permutations of 7 items by the permutations of the 5 identical process samples and 2 identical control samples.

Step 3: Calculate Possibilities for Distinct Process Samples

In part (b), the 5 process samples are different while the 2 control samples remain identical. Here, we first calculate the permutations of 5 distinct items, which is 5!. The total sequence length is still 7, so we use the formula \( \frac{7!}{2!} \) to find the number of possible sequences, as only the control samples are identical.

Step 4: Adjust for Control Sample at the Start

For part (c), the first sample must be a control sample. This leaves 6 positions to fill with 5 different process samples and 1 identical control sample. We first fix a control sample at the start, leaving us with 6 slots to fill with one more identical control sample and 5 distinct process samples, giving \( \frac{6!}{1!} \).

Step 5: Calculate Each Answer

For part (a), the number of sequences is \( \frac{7!}{5! \times 2!} = 21 \). For part (b), the number of sequences is \( \frac{7!}{2!} = 2520 \). For part (c), the number of sequences with the first sample fixed as a control is \( 5! = 120 \).

Key Concepts

Permutation with Identical Items

Permutations are all about arranging items in a sequence or order. One unique scenario in combinatorics is when you have identical items, where it doesn't matter how you swap them, they look exactly the same. For example, if you have five identical process samples and two identical control samples, the task is to determine how to sequence them for analysis in a laboratory setting. You can't simply use the basic permutation formula because that would count every arrangement, including ones that look the same due to identical items.

The specialized formula when dealing with identical items is\[\frac{n!}{n_1! \times n_2! \times \ldots \times n_k!}\]where \( n \) is the total number of items to arrange, and \( n_1, n_2, \ldots, n_k \) are the counts of each identical item group. Here, \( 7! \) represents the number of ways to arrange all samples if they were unique. The division by \( 5! \) and \( 2! \) is necessary to eliminate overcounting due to identical items. This method gives you an accurate number of distinct permutations. It's a neat trick when juggling identical items in combinatorics.

Sample Sequence Analysis

Sample sequence analysis involves determining all possible orders in which different samples can be processed. Imagine you have five distinct process samples and two identical control samples for a day of chemical analysis. The sequence matters because it affects calibration and outcomes.

While five distinct samples can be shuffled in any of the \( 5! \) ways, the two identical control samples reduce some of the variability since swapping them doesn't yield a new sequence. Therefore, the formula for calculating the total sequence possibilities transforms to \( \frac{7!}{2!} \). The 7! accounts for every potential order, while dividing by 2! removes the redundancy introduced by the identical control samples.

These types of calculations are critical in fields where accurate sequencing impacts the reliability of results. Whether you are working with data, chemicals, or other items, understanding how sequences of unique and identical items affect outcomes is key in sample analysis.

Control Sample in Experiments

In experimental settings, control samples are indispensable. They help ensure that instruments or processes are functioning correctly. In our exercise, a control sample is required at the beginning of testing each day. This stipulation alters the sequence possibilities.

By placing the control sample first, you effectively fill one slot in your sequence. The remaining sequence then consists of one more control sample and five distinct process samples. This reduces the problem to arranging six spaces, which is what \( 6!\) accounts for, to fill with 5 unique samples and a single additional control.

Implementing such constraints in your calculations is vital to ensure compliance with experimental protocols and to maintain the integrity of your results. Control samples help in validating the process, distinguishing between process-driven variations and genuine findings. Understanding how to correctly sequence them is as crucial as conducting the experiment itself.