Question

Let A, B and C be the sets such that $A\cup B=A\cup C$ and $A\cap B=A\cap C$. Show that B = C

Or

A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men and only three men got medals in all three sports? How many received medals in exactly two of the three sports?

Answer 1

We have $A\cup B=A\cup C$

$\Rightarrow (A\cup B)\cap C=(A\cup C)\cap C$

$\to (A\cap C)\cup (B\cap C)=C$

$\Rightarrow$ [$\because (A\cup C)\cap C=$and using distributive law $(A\cup B)\cap C=(A\cap C)\cup (B\cap C)]$

$\Rightarrow (A\cap B)\cup (B\cap C)=C[\because A\cap C=A\cap B]\dots \left(i\right)$

Again, $A\cup B=A\cup C$

$\Rightarrow (A\cup B)\cap B=(A\cup D)\cap B$

$\Rightarrow B=(A\cap B)\cup (C\cap B)$

$\Rightarrow$ [$\because (A\cup B)\cap B=B$ and using distributive law $(A\cup C)\cap B=(A\cap B)\cup (C\cap B)$]

$\Rightarrow B=(A\cap B)\cup (B\cap C)\dots \left(ii\right)$

From Eqs. (i) and (ii), we get.

B = C
or

Let F denotes the set of men who received medals in football, B denotes the set of men who received medals in basketball and C denotes the set of men who received medals in cricket.
Then, we have
$n\left(F\right)=38,n\left(B\right)=15.n\left(C\right)=20$

$n(F\cup B\cup c)=58andn(F\cap B\cap C)=3$

Now, $n(F\cup B\cup C)=n\left(F\right)+n\left(B\right)+n\left(C\right)-n(F\cap B)-n(8\cap C)-n(F\cap C)+n(F\cap B\cap C)$

$\Rightarrow 58=38+15+20-n(F\cap B)-n(B\cap C)-n(F\cap C)+3$

$\Rightarrow n(F\cap B)+n(B\cap C)+n(F\cap C)=76-58=18$

Now, number of men who received medals in exactly two of the three sports

$=n(F\cap B)+n(B\cap C)+n(F\cap C)-3n(F\cap B\cap C)=18-3\times 3=18-9=9$

Thus, 9 men received medals in exactly two of the three sports.