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Solutions to Berkeley's Math Circle: Combinatorics

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1

2^n-1 pwnage

2

3

444

4

5

There are 60 possibilities

6

There are 36 possibilities

7

8

308

10 kids x 31 flavors= 310 possibilities- 2 kids getting the same= 308


Nope, I'm pretty sure it's 31 x 31 x 31 x 31 x 31 x 31 x 31 x 31 x 31 x 31 (31 to the tenth power) minus 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 (the tenth pemutation of 31) It comes out to a somewhat gross number (in both senses of the word): 658, 683, 150, 600, 000 possibilities in all.

If you don't believe me, number tree it.

9

either 252 or 504 or non of these

10

11

12

There are 9 possible adjacent parking places if seven out of ten are already used.

13

14

15

16

17

18

19

20

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