A survey of 600 schools in India was conducted to gather information about their online teaching learning processes (OTLP). The following four facilities were studied.

F1: Own software for OTLP

F2: Trained teachers for OTLP

F3: Training materials for OTLP

F4: All students having Laptops

The following observations were summarized from the survey.

1. 80 schools did not have any of the four facilities – F1, F2, F3, F4.
2. 40 schools had all four facilities.
3. The number of schools with only F1, only F2, only F3, and only F4 was 25, 30, 26 and 20 respectively.
4. The number of schools with exactly three of the facilities was the same irrespective of which three were considered.
5. 313 schools had F2.
6. 26 schools had only F2 and F3 (but neither F1 nor F4).
7. Among the schools having F4, 24 had only F3, and 45 had only F2.
8. 162 schools had both F1 and F2.
9. The number of schools having F1 was the same as the number of schools having F4.

What was the total number of schools having exactly three of the four facilities?

The correct option is 200.

Let us break down the solution step-by-step using set theory and Venn diagrams for the four facilities: _F1, _F2, _F3, and _F4.
Let _N be the total number of schools surveyed. We are given:
$N=600$

Let us define the number of schools having a specific combination of facilities.
1. 80 schools did not have any of the four facilities. Therefore, the number of schools having at least one facility is:
$600\−80=520$

2. 40 schools had all four facilities:
$n(F_1\∩F_2\∩F_3\∩F_4)=40$

3. Let the number of schools with only _F1, only _F2, only _F3, and only _F4 be represented as:
$s_1=25$, $s_2=30$, $s_3=26$, and $s_4=20$.

4. The number of schools with exactly three of the facilities was the same irrespective of which three were considered.
Let this common value be $y$. Since there are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ ways to choose exactly three facilities from four, the total number of schools having exactly three of the four facilities is:
$4\⁢y$

5. 313 schools had _F2:
$n\left(F_2\right)=313$

6. 26 schools had only _F2 and _F3 (and neither _F1 nor _F4):
$n(\mathrm{only}\⁢F_2\⁢\&\⁢F_3)=26$

7. Among the schools having _F4, 24 had only _F3 (which means only _F3 and _F4), and 45 had only _F2 (which means only _F2 and _F4):
$n(\mathrm{only}\⁢F_3\⁢\&\⁢F_4)=24$
$n(\mathrm{only}\⁢F_2\⁢\&\⁢F_4)=45$

8. 162 schools had both _F1 and _F2:
$n(F_1\∩F_2)=162$

9. The number of schools having _F1 is equal to the number of schools having _F4:
$n\left(F_1\right)=n\left(F_4\right)$

Let us classify the schools having _F2 into disjoint categories based on the number of other facilities they have:
• Only _F2: 30
• Exactly two facilities including _F2:
- Only _F2 & _F1: let this be $x_{12}$
- Only _F2 & _F3: 26
- Only _F2 & _F4: 45
• Exactly three facilities including _F2:
- There are 3 combinations of three sets that contain _F2: (_F1, _F2, _F3), (_F1, _F2, _F4), and (_F2, _F3, _F4). Each of these has exactly $y$ schools. So, the contribution is $3\⁢y$.
• All four facilities: 40

Since the total number of schools having _F2 is 313, we can sum these components:
$30+(x_{12}+26+45)+3\⁢y+40=313$
Simplifying this equation:
$141+x_{12}+3\⁢y=313$
$x_{12}+3\⁢y=172$   — (Equation 1)

Now, let us use observation 8: 162 schools had both _F1 and _F2.
The region containing both _F1 and _F2 is composed of:
• Only _F1 & _F2: $x_{12}$
• Exactly three facilities including _F1 & _F2: (_F1, _F2, _F3) and (_F1, _F2, _F4), which equals $2\⁢y$.
• All four facilities: 40

Summing these gives:
$x_{12}+2\⁢y+40=162$
$x_{12}+2\⁢y=122$   — (Equation 2)

We can solve Equations 1 and 2 by subtracting Equation 2 from Equation 1:
$(x_{12}+3\⁢y)\−(x_{12}+2\⁢y)=172\−122$
$y=50$

The total number of schools having exactly three of the four facilities is:
$4\⁢y=4\×50=200$