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In a class of 35 students, 24 likes to play cricket, 5 likes to play both cricket and football. Find how many students like to play football?

Answer
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Hint:Put both the students in different sets according to their favourite sport and then apply the formula of a union of sets, which is given as:
$n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$

Complete step-by-step answer:
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.

Note:The union of two sets is again a set which contains the element which is in one of the two sets and the union of the two sets is expressed as $A \cup B$.
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 $A \cap B$.