Problem 32 - Algebra 2 and Trigonometry

Question

In a class of 22 students, the teacher calls on a student to give the answer to the first homework problem and then calls on a student to give the answer to the second homework problem. a. How many possible choices could the teacher have made if the same student was not called on twice? b. How many possible choices could the teacher have made if the same student may have been called on twice?

Step-by-Step Solution

Answer

a. 462 choices; b. 484 choices.

Step 1: Understanding Problem Part (a)

In part (a), we need to determine how many different ways the teacher can select two distinct students from the class of 22 to answer two different questions. This means no student can be selected more than once.

Step 2: Calculate First Choice for Part (a)

The teacher chooses the first student from the 22 available students. This gives us 22 possible choices for the first question.

Step 3: Calculate Second Choice for Part (a)

After selecting the first student, only 21 students remain for selection to answer the second question. This gives us 21 possible choices for the second question.

Step 4: Calculate Total for Part (a)

To find the total number of choices where no student is repeated, we multiply the number of choices for the first question by the number of choices for the second question. This is given by: \(22 \times 21 = 462\).

Step 5: Understanding Problem Part (b)

For part (b), the teacher can call on the same student for both questions. Therefore, the total number of choices is independent for each question.

Step 6: Calculate Total for Part (b)

The teacher selects a student from the 22 students for the first question and then again selects one from the same pool for the second question, leading to: \(22 \times 22 = 484\).

Key Concepts

Permutations

Permutations are fundamental in combinatorics. They involve arranging all or some elements in a specific sequence or order. When discussing permutations, we're interested in different sequences that can be formed by a set of elements. For instance, when arranging 3 students in line from a group, the order they stand in matters.

• Order matters
• Used when selecting without repetition
• Calculates distinct arrangements of a set

When solving permutation problems, imagine having slots to fill, with each slot representing a choice. If there are no repetitions, like in Part (a) of the exercise, once an element is chosen, it cannot be selected again. This means:
For the first slot, you have 22 choices. Once you pick a student, only 21 remain for the next slot. Thus, the permutations result in multiplying the options (e.g., 22 choices for the first question and 21 for the second, resulting in \(22 \times 21\)).

Combinations

Combinations are a key concept where the order of selection doesn't matter. This is a significant difference from permutations. If you select students for a group project and aim to choose them without caring about their order, you're dealing with combinations.

• Order does not matter
• Focuses on selecting a subset of items
• Used often in scenarios where grouping is emphasized over arrangement

In our exercise, however, combinations aren't directly applied, as we focus on individually distinct selections or repeating choices. Yet, knowing combinations is crucial because many real-world problems require understanding when order is insignificant.

Distinct Selections

Distinct selections refer to choosing elements in such a way that no element is repeated. It aligns with permutation principles where once an element is selected, it isn't available for another choice. This concept is essential in the original exercise, specifically Part (a).

• Elements are selected without repetition
• Each selection reduces available options
• Often used in sequential tasks

Applying this to our exercise, the teacher calls on students to answer questions where the same student cannot answer both. This requires distinct selections, highlighted by the multiplying sequence of possible choices: first, 22 options, then 21 possibilities they cannot re-select the first chosen student.

Repetitions Allowed

Allowing repetitions is a fascinating contrast to distinct selections. Here, each choice is made independently, meaning an element can be chosen multiple times. Repetition increases the total number of combinations significantly as seen in Part (b) of the exercise.

• Each choice is independent
• Elements can be reused
• Suitable where repetition is permitted

In our scenario, permitting the same student to answer both questions gives a selection process where each choice remains at 22 options per question. Therefore, 22 students may be selected for the first question and each student remains an option for subsequent questions, resulting in \(22 \times 22 = 484\) total possible sequences.