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MCIII Combo Help
Can I have hints on 4.30, 4.34, 4.35, 4.38, 4.39
For 4.36, I tried star and bars for each case where you have 0,1,2, and 3 empty boxes, and I got 29C20 + 28C20 27C20+ 26C20. What did I do wrong?
Thanks!
Here's some hints:
4.30: There are not that many arrangements. Can you "list" them out?
4.34: Consider complementary counting here. Hint: If one group has more than the other tow combined, can we say how many tickets they must have?
4.35: Consider cases based on how many colors are used. We can use the "up to rotation" to our advantage. For example, if all 4 colors are used, we can assume the "top" vertex of the tetrahedron is red.
4.38: Cases will be helpful to avoid overcounting 6s (there are many different ways to do this, one way is to consider cases based on the last 6, whether it's the ones, tens, hundreds, etc.) One advantage of this, if, for example, the leading digit is the only 6 is that then for the other digits there are 3 choices each that are 0, 1, and 2 mod 3. Try to use this to your advantage.
4.39: This one is quite tricky. A hint (still a little confusing, but try to think how this or something similar can help multiple times): By symmetry, $$\sum_{B\subset S} n(A\cap B) = \sum_{B\subset S} n(A\cap B^C).$$
For your question about 4.36, two things. One, be careful that you're ensuring that you have exactly 0, 1, 2, or 3 empty boxes (what does this tell us about the other boxes). Two, don't you also need to decide which boxes are empty?
Hope this helps!
I got:
4.30: 8
4.34: 73C71-(1C0+2C1+3C2+...+36C35)) = 1962
4.35:
1 color: 1*4
2 colors: 4C2*(1+1) = 12
3 colors: 4C3*1 = 4
4 colors: 2
to get 22.
4.36:
10C3*19C6+10C2+19C7+10C1+19C8+10C0*19C9
4.38: 12504
4.39 I still have no clue
4.38 looks good. For 4.36, some of the +'s should be *'s right? Looks like you have the right idea.
For 4.30, some are missing, answer should be 11.
For 4.34, for the complement you're subtracting, don't we need to multiply by 3 to choose which group it is as well?
For 4.35, be careful with choosing the colors. For example, with 2 colors, couldn't we color 2 vertices each the two colors or alternatively color 3 vertices one color and the last vertex a different color?
4.39 is quite tricky. In my opinion I would worry about it less than the others, but it does have a nice trick. Expanding on the hint from last time, we can calculate the inner sum as follows:$$\sum_{B\subset S} n(A\cap B) = \frac{1}{2}\sum_{B\subset S} \left(n(A\cap B)+n(A\cap B^C)\right)=\frac{1}{2}\sum_{B\subset S} n(A)=2^{n-1} n(A)$$(Here we use the symmetry mentioned in the last hint.) The key is that the symmetry allows us to get rid of the dependence on B.
Hope this helps!
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