The Pigeonhole Principle 
Vera has $10$ white socks, $10$ black socks, $10$ brown socks, $10$ blue socks, and $10$ red socks. How many socks (at a minimum) must she pull out of her sock drawer to ensure at least two matching pairs?
  (The two pairs cannot share a sock -- i.e., three socks of the same color don't count as forming multiple pairs.)

I thought, since we need $6$ socks to make at least one pair, we'd need $7$ to make at least two pairs, but that is incorrect. 
 A: 7 is incorrect because you can take
$3$ white, $1$ black, $1$ brown, $1$ blue, $1$ red.
You need at least $1$ more sock to make two pairs.
The correct answer is $8$: First, as you said if we choose $6$ socks we will have at least one pair. In fact we know something stronger, if we choose $7$ socks we either have $2$ pairs, or we have exactly $1$ pair and $1$ sock from each other color or $1$ pair and $3$ socks of the original color (just like mentioned above).
Now if you pick another sock, you either complete a match with the other colors or you have the situation described above ($3$ of one color and one of anything else).
At this point if you pick one more sock you have to either complete another pair of the same color as before, or complete another pair of another color. Either way you have two sets.
A: Hint:
What if Vera makes the following choice: White, Black , Brown, Blue, Red, White, White ...
A: Another argument why $8$ socks is sufficient: have Vera throw away one sock in each color of which she has an odd number of socks. Since she cannot have an odd number of socks in all colors (because then the total number of socks would be odd), she throws away at most $4$ socks and is left with at least $4$ socks. Furthermore, she now has an even number of socks in each color, so these at least $4$ socks must form at least $2$ pairs.
