# of ways

[westin]

$ of ways

Hi, the answer is 132 but I only got 126. can someone help what did I miss. thanks.

# of ways from Q to U = 4
# of ways from Q to U to E = 4 x 3 =12
# of ways from Q to U to E to U = 12/2 * 4 + 12/2 * 3 = 42 ( since 4 corners E cannot be reached at this stage, only 8 E left possible and half of the 8 E get to 4 U and half of 8 E get to 3 U)
# of ways from Q to U to E to U to E = 42 * 3 = 126

thanks!

 

[Otis]

Hi Westin. I'll post my paths, for you to compare. We may move Up, Down, Right or Left.

The non-Q entries are symmetrically placed around Q. Therefore, we can count just the possibilities that begin with moving up from Q and then multiply by 4.

Starting up, there are three choices for the second move (L, R and U).

Going left, without repeats, we have six paths:

U L D D
U L D L
U L L D
U L L U
U L U L
U L U R

With repeats, we have six more paths:

U L D U
U L L R
U L U D
U L R L
U L R U
U L R R

That's 12 paths total beginning with moves UL.

The second choice (after moving up from Q) is to go right. Due to symmetry, there are 12 paths beginning with UR.

The third choice is to move up. There are four such paths without repeats:

U U L D
U U L L
U U R D
U U R R

With repeats, we have five more paths:

U U L R
U U R L
U U D L
U U D R
U U D U

That's 9 paths total beginning with moves UU.

Therefore, we have a total of 12+12+9 paths beginning with U, and symmetry forms the same counts beginning with D, with L and with R.

33[imath]\times[/imath]4 = 132

?

 

[westin]

Thank you Otis. This is great help. I am now looking closely to why my initial approach is missing 6 ways. With your solution, it helps a lot. It seems that my approach should be similar to yours but I still don't get it what I am missing. It's getting late now, maybe I will take a look at it more closely tomorrow. If you can see what wrong with my initial logic, it will be great that you can give more more pointers. Again, thank you for the help!!!

 

[Otis]

I am now looking closely to why my initial approach is missing 6 ways ... I will take a look at it more closely tomorrow. If you can see what wrong with my initial logic, it will be great that you can give more more pointers.

Maybe you'd missed some moves with repeats, but after you look closely (comparing moves) I'm sure you'll see why your initial count is off.

By the way, I needed a lot more than five minutes to count em' (and I was confident with 132 only because you'd told us), but I had thought of drawing loops on the diagram partway through because I began to visualize the moves with non-repeats forming a type of clover leaves pattern, heh. Coming up with a nifty visualization would still have taken me well over five minutes.

 

[Cubist]

# of ways from Q to U = 4
# of ways from Q to U to E = 4 x 3 =12
# of ways from Q to U to E to U = 12/2 * 4 + 12/2 * 3 = 42

I think the last line in the quote above is wrong, you shouldn't divide by 2. Using @Otis 's method of considering the first move is up. Going from Q to U to E we end up on one of these three highlighted squares...

E U E U E     E U E U E     E U E U E
U E U E U     U E U E U     U E U E U
E U Q U E     E U Q U E     E U Q U E

Two of these options are surrounded by four U. But only one is surrounded by three U. Therefore...

# of ways from Q to U to E to U = 12*2/3 * 4 + 12*1/3 * 3 = 44
# of ways from Q to U to E to U to E = 44 * 3 = 132

 

[lookagain]

westin, is this another math contest problem or practice problem?

 

[westin]

Hi Cubist, i got it now, thank you. looks like I need to follow the path list otis instead of just counting to number of E that has three U or 4 U next to it. it looks clear now. but it is quite tricky.

yes, this is one of the questions in mathcount sprint round.

thank you all for help!