Cardinality of the Three Sets

Question 1

If $n(U) = 120$, $n(A) = 48$, $n(B) = 51$, $n(C) = 40$, $n(A \cap B) = 11$, $n(B \cap C) = 10$, $n(A \cap C) = 9$, and $n(A \cap B \cap C) = 4$ then find $n(A \cup B \cup C)$ and $n(\overline{A \cup B \cup C})$. Also show the information in the Venn-diagram.

Solution:

Finding $n(A \cup B \cup C)$:

For three sets, we use the inclusion-exclusion principle:

$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$

Substitute the given values:

$n(A \cup B \cup C) = 48 + 51 + 40 - 11 - 10 - 9 + 4$

Simplify step by step:

$n(A \cup B \cup C) = 139 - 30 + 4 = 113$

Finding $n(\overline{A \cup B \cup C})$:

The complement of $(A \cup B \cup C)$ represents elements not in any of the three sets:

$n(\overline{A \cup B \cup C}) = n(U) - n(A \cup B \cup C)$

Substitute the values:

$n(\overline{A \cup B \cup C}) = 120 - 113 = 7$

Verification:

Let's verify our calculations:

Sum of all regions: $32 + 34 + 25 + 7 + 6 + 5 + 4 + 7 = 120$ ✓

Check individual sets:

• $n(A) = 32 + 7 + 5 + 4 = 48$ ✓
• $n(B) = 34 + 7 + 6 + 4 = 51$ ✓
• $n(C) = 25 + 6 + 5 + 4 = 40$ ✓

Final Answers:

• $n(A \cup B \cup C) = 113$
• $n(\overline{A \cup B \cup C}) = 7$

Question 2

Among the 180 students who participated in SLC examination in 2071 from Nepal Madhyamik Vidhyaalaya, 86 passed in Science, 80 passed Maths and 76 passed in Nepali. Out of them, 26 passed in Science and Maths, 36 passed in Maths and Nepali as well as 32 passed in Science and Nepali but 20 did not pass all the subjects. Then,

1. Show the given information in the Venn-diagram.
2. Find the number of students who passed in all three subjects.

Solution:

Given information:

• Total students: $n(U) = 180$
• Passed Science: $n(S) = 86$
• Passed Maths: $n(M) = 80$
• Passed Nepali: $n(N) = 76$
• Passed Science and Maths: $n(S \cap M) = 26$
• Passed Maths and Nepali: $n(M \cap N) = 36$
• Passed Science and Nepali: $n(S \cap N) = 32$
• Did not pass all subjects: $n(\overline{S \cup M \cup N}) = 20$

a) Venn Diagram:

b) Finding students who passed all three subjects:

The statement "20 did not pass all the subjects" means 20 students passed none of the three subjects (failed completely).

Using the inclusion-exclusion principle for three sets:

$n(S \cup M \cup N) = n(S) + n(M) + n(N) - n(S \cap M) - n(M \cap N) - n(S \cap N) + n(S \cap M \cap N)$

Let $n(S \cap M \cap N) = x$. Substitute the given values:

$n(S \cup M \cup N) = 86 + 80 + 76 - 26 - 36 - 32 + x$ $n(S \cup M \cup N) = 242 - 94 + x = 148 + x$ -- (i)

Since 20 students passed none of the subjects, the number of students who passed at least one subject is:

$n(S \cup M \cup N) = n(U) - n(\overline{S \cup M \cup N}) = 180 - 20 = 160$ -- (ii)

Now we can solve for $x$ (from i & ii):

$160 = 148 + x$
so, $x = 160 - 148 = 12$

Therefore, 12 students passed in all three subjects.

Verification:

Students who passed at least one subject: $160$

Students who passed none of the subjects: $20$

Total: $160 + 20 = 180$ ✓

Question 3

A school distributed medals for the students in different events of a competition. 36 got medals in dance, 12 in drama and 18 in music. If only 45 students got medals and 4 students got medals in all three events, then find the number of students who got medals in exactly two events.

Solution:

Given information:

• Students who got medals in dance: $n(D) = 36$
• Students who got medals in drama: $n(Dr) = 12$
• Students who got medals in music: $n(M) = 18$
• Total students who got medals: $n(D \cup Dr \cup M) = 45$
• Students who got medals in all three events: $n(D \cap Dr \cap M) = 4$

Venn Diagram:

Step 1: Apply the inclusion-exclusion principle

For three sets:

$n(D \cup Dr \cup M) = n(D) + n(Dr) + n(M) - n(D \cap Dr) - n(Dr \cap M) - n(D \cap M) + n(D \cap Dr \cap M)$

Substitute the known values:

$45 = 36 + 12 + 18 - [n(D \cap Dr) + n(Dr \cap M) + n(D \cap M)] + 4$

Simplify:

$45 = 66 - [n(D \cap Dr) + n(Dr \cap M) + n(D \cap M)] + 4$ $45 = 70 - [n(D \cap Dr) + n(Dr \cap M) + n(D \cap M)]$

Solve for the sum of pairwise intersections:

$n(D \cap Dr) + n(Dr \cap M) + n(D \cap M) = 70 - 45 = 25$

Step 2: Find students in exactly two events

Students who got medals in exactly two events means:

• Students in D and Dr but not M: $n(D \cap Dr) - n(D \cap Dr \cap M)$
• Students in Dr and M but not D: $n(Dr \cap M) - n(D \cap Dr \cap M)$
• Students in D and M but not Dr: $n(D \cap M) - n(D \cap Dr \cap M)$

Total students in exactly two events:

$[n(D \cap Dr) - n(D \cap Dr \cap M)] + [n(Dr \cap M) - n(D \cap Dr \cap M)] + [n(D \cap M) - n(D \cap Dr \cap M)]$

Simplify:

$= [n(D \cap Dr) + n(Dr \cap M) + n(D \cap M)] - 3 \times n(D \cap Dr \cap M)$

Substitute the values:

$= 25 - 3 \times 4 = 25 - 12 = 13$

Therefore, 13 students got medals in exactly two events.

Verification:

Let's verify using individual regions:

Let:

• $a$ = students in exactly two events = 13
• $b$ = students in all three events = 4
• $c$ = students in exactly one event

Total students: $a + b + c = 45$

So: $13 + 4 + c = 45$

Therefore: $c = 28$ students got medals in exactly one event

Summary:

• Students in exactly one event: 28
• Students in exactly two events: 13
• Students in all three events: 4
• Total: $28 + 13 + 4 = 45$ ✓