Help me with this In a class of 30 students, 18 offer mathematics
(M), 15 offer science (S) and 13 offer english (€).
The number of students who offer all the three
subjects is equal to the number of students who
do not offer any of these subject. 10 students
offer both mathematics and english, 8 offer both
mathematics and science and 3 offers only
english and science. Determine: (i) the number
of students who offer all the three subjects (ii)
the number of students who offer only one
subject (iii) the number of learners who offer at
least two subject
To solve this problem, we can use the principle of inclusion-exclusion, which states that for three sets A, B, and C:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)
Given the information provided, let's calculate the answers to your questions step-by-step:
(i) The number of students who offer all three subjects:
According to the information provided, the number of students who offer all three subjects is equal to the number of students who do not offer any of these subjects. Let's denote this number as x.
n(A ∩ B ∩ C) = x
n(A' ∪ B' ∪ C') = x
Since offering all three subjects is the same as not offering none of the subjects, we can rewrite this equation as:
n((A ∩ B ∩ C)') = x
The symbol ' denotes the complement of a set. Therefore:
n(A' ∪ B' ∪ C') = x
We can now use the principle of inclusion-exclusion to calculate x:
x = n(A' ∪ B' ∪ C') = n(U) - n(A ∪ B ∪ C)
where n(U) is the total number of students, which is given as 30.
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)
n(A ∪ B ∪ C) = 18 + 15 + 13 - 10 - x - 8 + x
n(A ∪ B ∪ C) = 20
x = 30 - 20
x = 10
Therefore, the number of students who offer all three subjects is 10.
(ii) The number of students who offer only one subject:
We can calculate the number of students who offer only one subject by subtracting the number of students who offer multiple subjects from the total number of students.
n(Only one subject) = n(A) + n(B) + n(C) - 2 * n(A ∩ B) - 2 * n(A ∩ C) - 2 * n(B ∩ C) + n(A ∩ B ∩ C)
n(Only one subject) = 18 + 15 + 13 - 2 * 10 - 2 * x - 2 * 8 + x
n(Only one subject) = 15 - 4x
Substituting the value of x from the previous step:
n(Only one subject) = 15 - 4 * 10
n(Only one subject) = 15 - 40
n(Only one subject) = -25
Since a negative number of students is not possible, it means that there are no students who offer only one subject in this case.
(iii) The number of learners who offer at least two subjects:
We can calculate the number of students who offer at least two subjects by subtracting the number of students who do not offer any of these subjects from the total number of students.
n(At least two subjects) = n(A ∩ B) + n(A ∩ C) + n(B ∩ C) - n(A ∩ B ∩ C)
n(At least two subjects) = 10 + x + 3 - x
n(At least two subjects) = 10 + 3
n(At least two subjects) = 13
Therefore, the number of learners who offer at least two subjects is 13.
To solve this problem, it is helpful to use a Venn diagram or a table to organize the information.
Let's start by creating a Venn diagram:
1. Label the three circles as M (mathematics), S (science), and € (english).
2. Place the number 10 in the region where M and € overlap, since 10 students offer both mathematics and english.
3. Place the number 8 in the region where M and S overlap, since 8 students offer both mathematics and science.
4. Place the number 3 in the region where € and S overlap, since 3 students offer both english and science.
5. To find the number of students who offer only one subject, sum the number of students in each region where only one subject is present:
- M only: 18 - (10 + 8) = 0 students
- S only: 15 - (8 + 3) = 4 students
- € only: 13 - (10 + 3) = 0 students
Now, let's calculate the remaining values:
(i) The number of students who offer all three subjects:
- This is equal to the number of students in the intersection of all three circles.
- Since it is stated that the number of students who offer all three subjects is equal to the number of students who do not offer any of them, we can find this value by subtracting the total number of students who offer only one subject from the total number of students in the class:
- All three subjects: 30 - ((M only) + (S only) + (€ only)) - (No subject) = 30 - (0 + 4 + 0) - (No subject) = 26 students
(ii) The number of students who offer only one subject:
- This includes students who offer only mathematics, only science, or only english.
- The values for "(M only)" and "(€ only)" were already calculated and found to be 0.
- For "(S only)", we can calculate it by subtracting the number of students who offer both science and mathematics (8) and those who offer both science and english (3) from the total number of students who offer science (15):
- S only: 15 - (8 + 3) = 4 students
(iii) The number of learners who offer at least two subjects:
- This includes students who offer both mathematics and science (8) and those who offer both mathematics and english (10).
- To find this value, we sum those numbers:
- At least two subjects: (M and S) + (M and €) = 8 + 10 = 18 students
Therefore:
(i) The number of students who offer all three subjects is 26.
(ii) The number of students who offer only one subject is 4.
(iii) The number of learners who offer at least two subjects is 18.