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$n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$

It is given in the problem that there are 35 students in the class, out of which 24 likes to play cricket, 5 likes to play both cricket and football.

We have to find the number of students who like to play football.

Assume that the set $A$ defines the students who like to play cricket.

We know that there are 24 such students who like to play cricket, then

$n\left( A \right) = 24$

Also, assume that set $B$ defines the students who like to play football, and we have to find the number of such students,$n\left( B \right)$ who like to play football.

It is given in the problem that the total number of students are $35$, then it is denoted as:

$n(A \cup B) = 35$

We also know that the number of students who like both sports cricket and football are $5$, then it is expressed as:

\[n(A \cap B) = 5\]

Applying the formula of the union of sets which is given as:

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

Substitute the value of the given data:

$35 = 24 + n(B) - 5$

$35 - 24 + 5 = n(B)$

$n(B) = 16$

Therefore, there are 16 students who like to play football.

The intersection of two sets is again a set that contains all the elements that are in both the sets and it is expressed as $A \cap B$.