In a recently conducted scholarship exam, testbook has awarded 70 students. Out of which 38 students were awarded with tablet, 15 students were awarded with laptop, 20 students were awarded with smartphone, 5 students were awarded with tablet and laptop, 3 students were awarded with laptop and smartphone whereas 2 students were awarded with tablet and smartphone. Find the number of students who were awarded with all three.?

Option 2 : 7

Concept:

Let A, B and C be three finite sets and U is the finite universal set, then

1. n (A ∪ B) = n (A) + n (B) – n (A ∩ B)
2. n (A ∪ B) = n (A) + n (B) ⇔ A ∩ B = ϕ
3. n (A - B) = n (A) – n (A ∩ B) = n (A ∩ B’)
4. n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n(A ∩ B) – n (B ∩ C) – n (A ∩ C) + n (A ∩ B ∩ C)
5. n (A’ ∪ B’) = n = n (U) – n (A ∩ B)
6. n (A’ ∩ B’) = n = n (U) – n (A ∪ B)
7. n (A Δ B) = n (A) + n (B) – 2 n (A ∩ B)
8. n (A’) = n (U) – n (A)

Calculation:

Let, A = Number of students awarded with tablet, B = Number of students awarded with laptop and C = Number of students awarded with smartphone.

Given: n (A ∪ B ∪ C) = 70, n (A) = 38, n (B) = 15, n (C) = 20, n (A ∩ B) = 5, n (B ∩ C) = 3 and n (A ∩ C) = 2.

As we know that, if A, B and C are 3 finite sets, then

n (A ∪ B ∪ C) = n (A) + n (B) + n (C) – n(A ∩ B) – n (B ∩ C) – n (A ∩ C) + n (A ∩ B ∩ C)

⇒ 70 = 38 + 15 + 20 – 5 – 3 – 2 + n (A ∩ B ∩ C)

⇒ 70 = 63 + n (A ∩ B ∩ C)

⇒ n (A ∩ B ∩ C) = 7

Hence, there were 7 students who were awarded with all three.