1.1. The Westminster Math Club has 4 members: Ayanna, Beatrice, Charles,and Dante. How many different ways are there to choose a President and VicePresident for the club? List all possibilities. (For example, one possibilitywould be to have Ayanna as President and Dante as Vice President.)1.2. Make a table for the number of ways to choose a President and Vice-President for the club if the club has between 2 and 5 members. Note thatyour answer from Problem 1.1 will be one of the entries in this table.1.3. Plot your results from the table in Problem 1.2 on a graph, where thex-coordinate represents the number of members of the Westminster MathClub, and the y-coordinate represents the number of ways to choose thePresident and Vice-President. Be sure to label your points and horizontaland vertical axes.1.4. Make a table of points satisfying the equation y = x2 for x = 2, 3, 4,and 5.
Question
1.1. The Westminster Math Club has 4 members: Ayanna, Beatrice, Charles,and Dante. How many different ways are there to choose a President and VicePresident for the club? List all possibilities. (For example, one possibilitywould be to have Ayanna as President and Dante as Vice President.)1.2. Make a table for the number of ways to choose a President and Vice-President for the club if the club has between 2 and 5 members. Note thatyour answer from Problem 1.1 will be one of the entries in this table.1.3. Plot your results from the table in Problem 1.2 on a graph, where thex-coordinate represents the number of members of the Westminster MathClub, and the y-coordinate represents the number of ways to choose thePresident and Vice-President. Be sure to label your points and horizontaland vertical axes.1.4. Make a table of points satisfying the equation y = x2 for x = 2, 3, 4,and 5.
Solution 1
1.1. The number of ways to choose a President and Vice President from 4 members is calculated by permutation. The formula for permutation is P(n, r) = n! / (n-r)!, where n is the total number of items, and r is the number of items to choose. Here, n=4 (Ayanna, Beatrice, Charles, Dante) and r=2 (President and Vice President). So, P(4, 2) = 4! / (4-2)! = 12. The 12 different ways are: Ayanna-Beatrice, Ayanna-Charles, Ayanna-Dante, Beatrice-Ayanna, Beatrice-Charles, Beatrice-Dante, Charles-Ayanna, Charles-Beatrice, Charles-Dante, Dante-Ayanna, Dante-Beatrice, Dante-Charles.
1.2. The table for the number of ways to choose a President and Vice-President for the club if the club has between 2 and 5 members is:
| Members | Ways |
|---|---|
| 2 | 2 |
| 3 | 6 |
| 4 | 12 |
| 5 | 20 |
1.3. To plot the results, you would create a graph with the x-axis representing the number of members and the y-axis representing the number of ways to choose the President and Vice-President. Plot the points (2,2), (3,6), (4,12), and (5,20).
1.4. The table of points satisfying the equation y = x^2 for x = 2, 3, 4, and 5 is:
| x | y |
|---|---|
| 2 | 4 |
| 3 | 9 |
| 4 | 16 |
| 5 | 25 |
Solution 2
1.1. The number of ways to choose a President and Vice President from 4 members is calculated by permutation. The formula for permutation is nPr = n! / (n-r)!. Here, n is the total number of members and r is the number of positions to fill. In this case, n=4 and r=2. So, the number of ways is 4P2 = 4! / (4-2)! = 12. The possibilities are: Ayanna-Beatrice, Ayanna-Charles, Ayanna-Dante, Beatrice-Ayanna, Beatrice-Charles, Beatrice-Dante, Charles-Ayanna, Charles-Beatrice, Charles-Dante, Dante-Ayanna, Dante-Beatrice, Dante-Charles.
1.2. The table for the number of ways to choose a President and Vice-President for the club if the club has between 2 and 5 members is:
| Members | Ways |
|---|---|
| 2 | 2 |
| 3 | 6 |
| 4 | 12 |
| 5 | 20 |
1.3. To plot the results, you would create a graph with the number of members on the x-axis and the number of ways on the y-axis. Plot the points (2,2), (3,6), (4,12), and (5,20).
1.4. The table of points satisfying the equation y = x^2 for x = 2, 3, 4, and 5 is:
| x | y |
|---|---|
| 2 | 4 |
| 3 | 9 |
| 4 | 16 |
| 5 | 25 |
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