[Prof. Templeton]

How many ways can you form the sentence "Rise to vote sir" from this configuration. You may go forward and backward, up and down, but each sucessive letter must adjoin one another.

						R						


					R	I	R					


				R	I	S	I	R				


			R	I	S	E	S	I	R			


		R	I	S	E	T	E	S	I	R		


	R	I	S	E	T	O	T	E	S	I	R	


R	I	S	E	T	O	V	O	T	E	S	I	R


	R	I	S	E	T	O	T	E	S	I	R	


		R	I	S	E	T	E	S	I	R		


			R	I	S	E	S	I	R			


				R	I	S	I	R				


					R	I	R					


						R						

[/codebox]

[Guest]

If you cannot go diagonal..

24???

[Guest]

52

Edited January 19, 2009 by solarfish

[Guest]

4096 (or 2^12) ... the more I think of this, the shakier my logic stands, so I@ll leave it as an uneducated guess

[Guest]

Can the same letter be used twice?

R I R

R I S I R

R I S E S I R

[color="#FF0000"][b]R[/b] I S E T[/color] E S I R

R [color="#FF0000"]I[/color] S E T [color="#FF0000"]O[/color] T E S I R

R I [color="#FF0000"]S E T O V[/color] O T E S I R

R I S E T O T E S I R

R I S E T E S I R

R I S E S I R

R I S I R

R I R

R

						R						

[Guest]

How many ways can you form the sentence "Rise to vote sir" from this configuration. You may go forward and backward, up and down, but each sucessive letter must adjoin one another.

Somewhere in the neighborhood of 2¹⁵

Edited January 19, 2009 by d3k3

[Prof. Templeton]

Somewhere in the neighborhood of 2¹⁵

to 2¹⁶, but not quite.

[Guest]

49152, but I don't think that's the correct answer.

[Guest]

36864

[Guest]

3^12

[Guest]

to 2¹⁶, but not quite.

I missed the bit where you can turn around at "V" and go back from whence you came, which would make it more like 64,516. Unless you can't use the same letter twice...

[Guest]

One more thought:

For example, considering each letter separately such as

R=24, because there are 24 initial Rs you could start from.

Then I= 2(20/24) + 1(4/24), because for 20 of the selected Rs you would have 2 Is to move to, but for 4 of the Rs you would only have one I to move to.

Then S= 2(16/20) + 1(4/20), because for 16 of the selected Is you would have 2 Ss to move to, but for 4 of the Is you would only have one S to move to.

and you continue... (I'll write all the letters out)

E= 2(12/16) + 1(4/16)

T= 2(8/12) + 1(4/12)

O= 2(4/8) + 1(4/8)

V= 1

O= 4

T= 3

E= 3(4/8) + 2(4/8), because for 4 of the Ts you would have 3 Es to select, but for the other 4 Ts you would only have 2 Es to select.

S= 3(4/12) + 2(8/12)

I= 3(4/16) + 2(12/16)

R= 3(4/20) + 2(16/20)

And when you multiply every letter together you get........ (assuming I typed it into my calculator correctly).......... 120,062.25

Amazing. I highly doubt that the answer is not an integer.

So....120,063 paths? Perhaps?

If this is wrong, could someone please explain where my logic is skewed?

[Guest]

63,504

Start with the top R. Call it R1. Going down to the right we have R2, R3, and R4.

Every single valid path must pass through V, of which there is only one at the center. So this is really a problem of how many different paths are there from the V to a given R.

R1 may only be reached one way. R2 may be reached six ways, R3 fifteen ways, and R4 twenty ways. By symmetry you can apply these numbers all around the figure.

There are four R1's, eight each of R2 and R3, and four R4's.

Therefore there are (4 x 1) + (8 x 6) + (8 x 15) + (4 x 20) = 252 valid paths from the V to an R. Similarly, there are 252 valid paths to the V from an R. 252 x 252 = 63,504 paths.

[Guest]

Start with the top R. Call it R1. Going down to the right we have R2, R3, and R4.

Every single valid path must pass through V, of which there is only one at the center. So this is really a problem of how many different paths are there from the V to a given R.

R1 may only be reached one way. R2 may be reached six ways, R3 fifteen ways, and R4 twenty ways. By symmetry you can apply these numbers all around the figure.

There are four R1's, eight each of R2 and R3, and four R4's.

Therefore there are (4 x 1) + (8 x 6) + (8 x 15) + (4 x 20) = 252 valid paths from the V to an R. Similarly, there are 252 valid paths to the V from an R. 252 x 252 = 63,504 paths.

63,504

63,504 is correct.

[Guest]

if the phrase should only be formed in a straight line. Is it valid if the phrase is formed in an "L" shape?

If yes, then we have (7*7)*2 number of ways.

7*7: for each row and column

*2: because each pharse can also be formed backwards, so that if you formed one pharse..you can go backwards and form another.

Am I correct?

[Guest]

if the phrase should only be formed in a straight line. Is it valid if the phrase is formed in an "L" shape?

If yes, then we have (7*7)*2 number of ways.

7*7: for each row and column

*2: because each pharse can also be formed backwards, so that if you formed one pharse..you can go backwards and form another.

Am I correct?

[Guest]

Oops..my answer got posted twice...Seems everybody else have far larger number than mine. Will have to recalculate if the same letter can be used twice?

[Prof. Templeton]

Start with the top R. Call it R1. Going down to the right we have R2, R3, and R4.

Every single valid path must pass through V, of which there is only one at the center. So this is really a problem of how many different paths are there from the V to a given R.

R1 may only be reached one way. R2 may be reached six ways, R3 fifteen ways, and R4 twenty ways. By symmetry you can apply these numbers all around the figure.

There are four R1's, eight each of R2 and R3, and four R4's.

Therefore there are (4 x 1) + (8 x 6) + (8 x 15) + (4 x 20) = 252 valid paths from the V to an R. Similarly, there are 252 valid paths to the V from an R. 252 x 252 = 63,504 paths.

63,504

That's it!

63,504 is correct.

Correct.

When the sentence contains 2n+1 letters (13 in this case so n=6) then the formula for the number of different ways is [4(2ⁿ-1)]².

[Guest]

My proposed solution uses two observations:

1) All paths must go through the V in the center.

2) There are an equal number of paths to (top) and from (below) the V

Begin by counting the number of paths from each starting point to the V:

1

6

9

10

6

1

The top right side is the same. So we add these all up to get 67 pathways to the V. It's easy to see that there will also be 67 pathways from the V. The total number of combinations will be 67x67 = 4489

There's probably a more elegant solution out there. But this only took a minute to get.