There are 35 students participating in school competition. Each student must take at least one of these three competitions. 20 students participate in Cricket, 15 students particip... There are 35 students participating in school competition. Each student must take at least one of these three competitions. 20 students participate in Cricket, 15 students participate in basketball, and 13 students participate in football. If 9 students participate in at least two competitions, then what is the number of students who participate in all the competitions? Read more

Steps to Solve

1. Identify Given Information

We know the following:

• Total number of students, $N = 35$
• Students in Cricket, $C = 20$
• Students in Basketball, $B = 15$
• Students in Football, $F = 13$
• Students participating in at least two competitions, $A = 9$

2. Set Up the Equation for Total Students

Using the principle of inclusion-exclusion for three sets, the equation can be written as:

$$ N = C + B + F - (AB + AF + BF) + X $$

Here, $AB$, $AF$, and $BF$ are the number of students participating in two competitions, and $X$ is the number of students participating in all three competitions.

3. Express the Groups in Terms of Known Values

Since $A$ (the number of students in at least two competitions) includes students in all pairs of competitions plus those in all three, we also have:

$$ A = AB + AF + BF - 2X $$

Substituting $A = 9$, we get:

$$ 9 = AB + AF + BF - 2X $$

4. Substitute Known Values Into Equations

Now we can substitute into the first equation:

$$ 35 = 20 + 15 + 13 - (AB + AF + BF) + X $$

This simplifies to:

$$ 35 = 48 - (AB + AF + BF) + X $$

5. Rearranging Equations

Rearranging gives us:

$$ AB + AF + BF - X = 13 $$

From the second equation, we substitute $AB + AF + BF = 9 + 2X$ into the arrangement:

$$ (9 + 2X) - X = 13 \implies 9 + X = 13 $$

6. Solve for $X$

Now we solve for $X$:

$$ X = 13 - 9 \implies X = 4 $$

Thus, the number of students participating in all competitions is 4.