I'll use subscript notation for brevity, i.e. 
.
By the chain rule,
We have
and
When 
, we have
and the partial derivatives take on values of
So we end up with
Answer:
5
Step-by-step explanation:
The 32 that have blue and green ribbons include the 16 that have all three, so there are only 32 -16 = 16 that have only blue and green ribbons.
The 31 that have green and white ribbons likewise include the 16 with all three, so there are only 31 -16 = 15 that have only green and white ribbons.
The 38 that have blue and white ribbons include the 16 with all three, so there are only 38 -16 = 22 that have only green and white ribbons.
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If we add the numbers of blue, green, and white ribbons, we are counting twice the numbers that have 2 ribbons, and 3 times the numbers that have 3 ribbons. We want to count each kind of ribbon-holder only once. Hence the number of individual dogs with any number of ribbons is only ...
62 +55 +63 -(16 +15 +22) -2(16) = 95
Of the 100 dogs, 95 have ribbons, so 5 dogs have not learned any tricks.
