In a competition with 80 participants, a school awarded medals in different categories. 36 medals in dance (D), 12 medals in dramatics (R) and 18 medals in music (M). If these medals went to a total of 45 persons and only 4 persons got medals in all the three categories, find the number of persons who received medals in only one of the categories.
From the problem we have
$nD=36,nR=12,nM=18, n(D\cup{R}\cup{M})=45,n(D\cap{R}\cap{M})=4$
and let $D\cap{M}=y+4, D\cap{R}=x+4,R\cap{M}=z+4$
and by inclusion/exclusion,i.e.
$n(D\cup{R}\cup{M})=nD+nR+nM-n(D\cap{M})-n(D\cap{R})-n(R\cap{M})+n(D\cap{R}\cap{M})$
$45=36+12+18-(y+4)-(x+4)-(z+4)+4$ which simplifies into $13=x+y+z$
Let $a$ be the number of persons who recieves medals in Dance $(D)$ only. Similarly, let $b,c$ be the number of person who receives medals only in Dramatics$(R)$ and Music $(M)$ respectively.
Then,
using $13=x+y+z$
$a=32-(x+y)\sim{a}=z+19$
$b=8-(x+z)\sim{b}=y-5$
$c=14-(y+z)\sim{c}=x+1$
adding the equations we get
$a+b+c=15+(x+y+z)\sim{a+b+c}=15+13=28$
Therefore, the number of persons who received medals in only one of the categories is 28
However, I want to find further the value of $x,y,z$ and $a,b,c$ is it possible given such information?
No.
There are multiple solutions to these constraints. For example:
$a=20, b=4, c=4, x=3, y=9, z=1$ works
But $a=22, b=3, c=3, x=2, y=8, z=3$ also works
And many more ...