Solutions
1) 12 socks. In the worst case, I get all 10 black socks and the last two are both blue.
2) 3 socks. Picking 2 doesn't guarantee a pair, as there could be one of each colour. The third sock has to match one of the colours and make a pair, though.
3) 11 socks. In the worst case, I get all 10 black socks and the last one is blue.
Solutions
1) 19 gloves. Worst case is that I pick all the green and yellow gloves. The next two gloves must both be red.
2) 6 gloves. The first 3 could all be different. The fourth will match one of them to make a pair. The worst case scenario now is that the fifth glove is also that colour, which doesn't have a pair. The sixth then must match one of the other gloves to make a second pair.
3) 19 gloves. I could pick up all 7 yellow gloves first. To guarantee that I have 8 green gloves, I need to grab another 12 gloves because I might pick all the red gloves first.
Solution
18 socks. At a glance, this looks like there could be some algebra coming up, but no. If I pick 18 socks, I could pick up all 16 blue socks and then the next two must both be black.
Discussion point - An assumption was made to answer that question. What if \(n=1\)?
Solution
\(n+2\) socks. If I pick \(n+2\) socks, I could pick up all \(n\) black socks and
then the next two must both be blue.
Discussion point - An assumption was made again! What if \(n=1\)? Or 0?
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