# 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.

• The 7th can be the same color as the first pair ... – Bram28 Apr 29 '18 at 14:40
• If you had only two of each coloured sock it would work. – Nelson Apr 30 '18 at 1:23

Hint:

What if Vera makes the following choice: White, Black , Brown, Blue, Red, White, White ...

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.

• Can't Vega choose 10 white socks, 1 Black, 1 Brown, 1 Blue, 1 Red ? – onurcanbektas Apr 29 '18 at 14:41
• @onurcanbektas 4 white socks count as 2 pairs. (so 10 counts as 5...) – Yanko Apr 29 '18 at 14:42
• Yeah, I missed that, thanks for pointing out. – onurcanbektas Apr 29 '18 at 14:43
• "if we choose 6 socks we ... either have 2 pairs, or we have exactly 1 pair and 1 sock from each other color." this is incorrect. we can have 3,1,1,1,0 of each color which gives us only 1 pair, since socks can't be shared among pairs. we need 7 for that (with 3,1,1,1,1 the worst configuration). – Will Ness Apr 29 '18 at 21:48
• @WillNess Thanks for the correction. – Yanko May 1 '18 at 7:58

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.