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
We will draw what is called a "Venn diagram":
First draw a big rectangle to contain all 24 boxes.
Now in that rectangle draw a big circle for the 13 boxes that
contain nickels and label it N:
Next draw another big circle overlapping it to contain the 8 boxes
that contain dimes and label it 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
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:
and in the main drawing:
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:
and in the main drawing:
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:
` `
and in the main drawing:
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:
So the answer is 8.
Edwin