1. There are 18 chairs numbered from 1 through 18 around a circular table. How many ways can three people be seated, so that no two people are adjacent?
Note: the answer is NOT 3240 or 180.
2. Jeff the fly starts at the point (0,0) in the coordinate plane. At each step, Jeff takes a step to the right, left, up, or down. After steps, how many different points could Jeff end up at?
Note: the answer is NOT 286 or 40.
PLEASE!! THANK YOU!!
1) There are 18*17*16 ways to choose the three seats, if people can be adjacent. We must subtract the cases where there are two people adjacent, which is 18*13*6. So the number of ways is 18*17*16 - 18*13*6 = 3492.
2) Jeff can reach any point inside the square with vertices (10,10), (10,-10), (-10,-10), and (-10,10), so the answer is 21^2 = 441.
+865 1) Without any restrictions, there are 18*17*16 = 4896 ways that three people can be seated. Then, we have to subtract the cases where they are adjacent.
Case 1: Where all three people are next to each other in a "boomerang" shape. There are 18 ways to choose three consecutive seats. There are then 3! = 6 ways to place those three people, which gives us 18*6=108 possible seatings where they are next to each other.
Case 2: There are two people sitting next to each other, and the third person is not next to either of them (loner, like me).
There are 18 ways to choose two consecutive seats for the two people. There are then 14 ways to choose the third seat for the loner. After the three seats have been chosen, there are 3!=6 ways to place the three people, which gives us 18*14*6=1512 possible seatings.
Subtracting all of them: 4896-1512-108=3276 ways.
Hope this is right!
2) Can't help you since there is a typo. "After steps". No number.
Thank you so much! Your answer to #1 is correct. I apologize for the typo, the number is 10
+865 In that case,
The ending point of the path Jeff made depends on all the moves Jeff made, not the order. Consider a random 10-step path Jeff could take. NESWNNESEW (N for North, S for south, W for west, E for East. Reduce our path by cancelling each E-W and each N-S pair, which will not change our endpoint. So our example reduces to NE. Notice that our reduced path must have an even number of moves, and it is thus possible to end at any point (x,y) where x+y is even and |x| + |y| is no greater than 10.'
If |x| + |y| = 0, there is one possibility for (x,y).
If |x| + |y| <= (less than equal to) 2, there are 9 possible paths ((0,0) (-+1,-+1) (0, +-2) (+-2,0)); these solutions will form a 3x3 diagonal grid
The solutions with |x| + |y| <= 10 will form an 11x11 grid, so there are 11^2=121 ways
Sorry if I didn't explain this very well, I'm in a bit of a rush.
