Question 265920
My entire nickel and dime collection is housed in my cardboard box collection, which consists of exactly 24 boxes. 13 of the boxes contain nickels, 8 contain dimes and 5 contain both nickels and dimes. How many of the boxes contain neither

<pre><font size = 4 color = "indigo"><b>

We will draw what is called a "Venn diagram":

First draw a big rectangle to contain all 24 boxes.
 
{{{drawing(400,400,-3.2,3.2,-3.2,3.2, rectangle(-3,-2,3,2) )}}}
 
Now in that rectangle draw a big circle for the 13 boxes that
contain nickels and label it N:
 
{{{drawing(400,400,-4,4,-4,4, rectangle(-3.9,-2.9,3.9,2.9),
circle(-1,0,2), locate(-2.3,2,N)   )}}}
 
Next draw another big circle overlapping it to contain the 8 boxes
that contain dimes and label it D:
 
{{{drawing(400,400,-4,4,-4,4, rectangle(-3.9,-2.9,3.9,2.9),
circle(-1,0,2), circle(1,0,2), locate(-2.3,2,N), locate(2.3,2,D)  )}}}
 
Now the 5 boxes which contain both nickels and dimes will have to go
in the overlapping part of those two circles.  So we'll write {{{5}}} 
in there to represent the 5 boxes in that region.  Those 5 have to
be in BOTH CIRCLES at the same time! The way for those 5 to be in both
circles at the same time is for them to be in the overlapping part of the
two circles, which is football-shaped:

{{{drawing(100,173,-1,1,-1.1,1.2, rectangle(-3.9,-2.9,3.9,2.9),
circle(-1,0,2), circle(1,0,2), locate(-2.3,2,N), locate(2.3,2,D),
locate(0,0,5) )}}} 

and in the main drawing:
 
{{{drawing(400,400,-4,4,-4,4, rectangle(-3.9,-2.9,3.9,2.9),
circle(-1,0,2), circle(1,0,2), locate(-2.3,2,N), locate(2.3,2,D),
 locate(0,0,5)
 
  )}}}
 
Since there are 13 boxes which contain nickels, the rest of those 13 
boxes besides the 5 that contain both nickels and dimes, that is, 
13-5, or 8, will have to go in this region, which is moon-shaped,
in which we write 8:
 
{{{drawing(200,400,-4,-.1,-4,4, locate(-2,0,8),
circle(-1,0,2), circle(1,0,2)  )}}}  
 
and in the main drawing:
 
{{{drawing(400,400,-4,4,-4,4, rectangle(-3.9,-2.9,3.9,2.9),
circle(-1,0,2), circle(1,0,2), locate(-2.3,2,N), locate(2.3,2,D),
locate(0,0,5), locate(-2,0,8))}}}

Notice that the circle labeled N now contains the 13 boxes which
contain nickels, 8 of them in the left moon-shaped part of the
circle and 5 in the football-shaped part of the circle.  The 8
contain nickels only and the 5 contain both nickels and dimes. 

Since there are 8 boxes which contain dimes, the rest of those 8 
besides the 5 that are in both circles, that is, 8-5, or 3, will 
be in the right moon-shaped region, in which we write 3:

`                      `{{{drawing(200,400,0,4,-4,4, locate(2,0,3),
circle(-1,0,2), circle(1,0,2)  )}}}  
 
and in the main drawing:
 
{{{drawing(400,400,-4,4,-4,4, rectangle(-3.9,-2.9,3.9,2.9),
circle(-1,0,2), circle(1,0,2), locate(-2.3,2,N), locate(2.3,2,D),
locate(0,0,5), locate(-2,0,8), locate(2,0,3)
 
  )}}}
 
Notice that the circle labeled D now contains the 8 boxes which
contain dimess, 3 of them in the right moon-shaped part of the
circle and 5 in the football-shaped part of the circle.  The 3
contain dimes only and the 5 contain both nickels and dimes. 

Now we have accounted for 8 + 5 + 3 or 16 of the 24 boxes.  That
leaves 24-16, or 8 boxes which contain neither nickels nor dimes,
and so those 8 are outside the two circles, like this:

{{{drawing(400,400,-4,4,-4,4, rectangle(-3.9,-2.9,3.9,2.9),
circle(-1,0,2), circle(1,0,2), locate(-2.3,2,N), locate(2.3,2,D),
locate(0,0,5), locate(-2,0,8), locate(2,0,3), locate(-3,-2,8)
 
  )}}}

So the answer is 8.
 
Edwin</pre>