We are given the following information about students in a Faculty of Science: - Total number of students = 1000 - Number of students who offered Mathematics = 360 - Number of students who offered Chemistry = 420 - Number of students who offered Biology = 320 - Number of students who offered Mathematics and Chemistry = 140 - Number of students who offered Mathematics and Biology = 160 - Number of students who offered Chemistry and Biology = 180 - Number of students who offered all three subjects = 100 We need to find: 17. The number of students that offered Mathematics but offered neither Chemistry nor Biology. 18. The number of students who did not offer any of the three subject combinations. 19. The number of students who offer at least two subjects.

1. Problem Description

We are given the following information about students in a Faculty of Science:

- Total number of students = 1000

- Number of students who offered Mathematics = 360

- Number of students who offered Chemistry = 420

- Number of students who offered Biology = 320

- Number of students who offered Mathematics and Chemistry = 140

- Number of students who offered Mathematics and Biology = 160

- Number of students who offered Chemistry and Biology = 180

- Number of students who offered all three subjects = 100

We need to find:

1

7. The number of students that offered Mathematics but offered neither Chemistry nor Biology.

1

8. The number of students who did not offer any of the three subject combinations.

1

9. The number of students who offer at least two subjects.

2. Solution Steps

Let M, C, and B represent the sets of students offering Mathematics, Chemistry, and Biology, respectively.

We are given the following:

$|U| = 1000$

$|M| = 360$

$|C| = 420$

$|B| = 320$

$|M \cap C| = 140$

$|M \cap B| = 160$

$|C \cap B| = 180$

$|M \cap C \cap B| = 100$

1

7. To find the number of students who offered Mathematics but neither Chemistry nor Biology, we need to find $|M \cap C' \cap B'|$.

$|M \cap C' \cap B'| = |M| - |M \cap (C \cup B)|$

$|M \cap (C \cup B)| = |(M \cap C) \cup (M \cap B)| = |M \cap C| + |M \cap B| - |M \cap C \cap B| = 140 + 160 - 100 = 200$

So, $|M \cap C' \cap B'| = |M| - |M \cap (C \cup B)| = 360 - 200 = 160$

1

8. To find the number of students who did not offer any of the three subjects, we need to find $|(M \cup C \cup B)'|$.

$|M \cup C \cup B| = |M| + |C| + |B| - |M \cap C| - |M \cap B| - |C \cap B| + |M \cap C \cap B|$

$|M \cup C \cup B| = 360 + 420 + 320 - 140 - 160 - 180 + 100 = 1200 - 480 + 100 = 820$

So, $|(M \cup C \cup B)'| = |U| - |M \cup C \cup B| = 1000 - 820 = 180$

However, it appears the problem has a typo, as the number of students who did not take any of the three is calculated as 180 not 440

1

9. To find the number of students who offer at least two subjects, we need to find $|(M \cap C) \cup (M \cap B) \cup (C \cap B)|$. This is equivalent to finding $|M \cap C| + |M \cap B| + |C \cap B| - 2|M \cap C \cap B|$

$|M \cap C| + |M \cap B| + |C \cap B| - 2|M \cap C \cap B| = 140 + 160 + 180 - 2*100 = 480 - 200 = 280$

3. Final Answer

1

7. The number of students that offered Mathematics but offered neither Chemistry nor Biology is

1

6

0.

1

8. The number of students who did not offer any of the three subject combinations is

1

8

0.

1

9. The number of students who offer at least two subjects is 280.