Question

There are 19 hockey players in a club. On a particular day 14 were wearing the prescribed hockey shirts, while 11 are wearing the prescribed hockey pants. None of them were without hockey pants or hockey shirts. How many of them were in complete hockey uniform?

Answer 1

Hint – This is problems related to sets, So let A be the event related to shirts and B be the event related to pants, now $n\left( {A \cup B} \right) = 19$ and $n\left( A \right) = 14,n\left( B \right) = 11$, so use the direct formula for $n\left( {A \cup B} \right)$which is $n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$. This will help finding the value of $n\left( {A \cap B} \right)$.

Complete step-by-step answer:
Total hockey players = 19.
Now it is given that on a particular day 14 were wearing the prescribed hockey shirts.
And 11 were wearing the prescribed hockey pants.
It is also given that none of them was without hockey pants and hockey shirts.
So the above information should be completely satisfying.
Let A = shirts and B = pants.
So according to above information
$n\left( {A \cup B} \right) = 19$, (union of A and B)
And $n\left( A \right) = 14,n\left( B \right) = 11$
So according to set relation we have,
$\Rightarrow n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$
Where $n\left( {A \cap B} \right)$ are the intersection of A and B i.e. out of 19 players how many are in complete hockey uniform.
Now substitute the values we have,
$\Rightarrow 19 = 14 + 11 - n\left( {A \cap B} \right)$
$\Rightarrow n\left( {A \cap B} \right) = 14 + 11 - 19 = 25 - 19 = 6$
So this is the number of hockey players who were in complete hockey uniform.
So this is the required answer.
Hence option (B) is the required answer.

Note – There can be another way of solving this problem via direct approach of vein diagram, as we can see in the figure, one left circle corresponds to event A and one right circle corresponds to event B, the complete two circles represents $n\left( {A \cup B} \right)$ and the intersection region of the two circles represents $n\left( {A \cap B} \right)$. Clearly the intersection region comes twice if once the complete left A region circle is considered and one complete right region B is considered therefore we have reduced this region once, to get the complete $n\left( {A \cup B} \right)$.