Beginning with letter A at the top of the triangle and reading down,
always passing from a letter to an adjoining letter, how many ways
is it possible to read abracadabra ?
-bapcha-
For each step, you always have two possible choices: left or right.
So the answer must be 2^10 = 1024.
That was trivial. What about this one:
A A A A A A A A A A A
B B B B B B B B B B B
R R R R R R R R R R R
A A A A A A A A A A A
C C C C C C C C C C C
A A A A A A A A A A A
D D D D D D D D D D D
A A A A A A A A A A A
B B B B B B B B B B B
R R R R R R R R R R R
A A A A A A A A A A A
The rules are the same, and you may also choose where to start at the top.
(No, I don't know the answer.)
Bj|rn Munch ("The Man With a Pipe in His Name") bjo...@idt.unit.no
===========
1 1
2 1 1
3 1 2 1
4 1 3 3 1
5 1 4 6 4 1
6 1 5 10 10 5 1
7 1 6 15 20 15 6 1
8 1 7 21 35 35 21 7 1
9 1 8 28 56 70 56 28 8 1
10 1 9 36 84 126 126 84 36 9 1
11 1+10+45+120+210+252+210+120+45+10+1 = 1024 = 2**(11-1) = 2**(N-1)
where N = number of rows
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-- PUZZLE DELETED --
It reminds me of a cute little puzzle that an ex-math-teacher told me.
(It was the only thing I learnt from her (sorry Els))
The rules are the same, start at top, end at bottom.
A
B B
R R R
A A A A
C C C C C
A A A A A A
D D D D D
A A A A
B B B
R R
A
If you are able to solve this diamond, you will be able to solve
any other puzzle of this type (It's just a trick).
Izak van Langevelde.
1 A 1 1 1 1 1 1 1 1 1 1 1
2 B 2 2 2 2 2 2 2 2 2 2 1
3 R 2 4 4 4 4 4 4 4 4 4 3
4 A 6 8 8 8 8 8 8 8 8 7 3
5 C 6 14 16 16 16 16 16 16 16 15 10
6 A 20 30 32 32 32 32 32 32 31 25 10
7 D 20 50 62 64 64 64 64 64 63 56 35
8 A 70 112 126 128 128 128 128 127 119 91 35
9 B 70 182 238 254 256 256 256 255 246 200 126
10R 252 420 492 510 512 512 511 501 446 326 126
11A 252 +672 +912 +1002+1022+1024+1023+1012+947 +772 +452 = 9090
In article <65...@star.cs.vu.nl> iav...@cs.vu.nl (Langevelde van IA) writes:
>
>In article <1990May15.1...@idt.unit.no> bjo...@idt.unit.no (Bj|rn Munch) writes:
>>
>>In article <6...@lee.SEAS.UCLA.EDU>, bap...@edison.seas.ucla.edu (Bapcha)
>>writes:
>>|>
[ stuff deleted ]
>>|>
>>|> Beginning with letter A at the top of the triangle and reading down,
>>|> always passing from a letter to an adjoining letter, how many ways
>>|> is it possible to read abracadabra ?
>>|>
>>|> -bapcha-
>>
>
>
>It reminds me of a cute little puzzle that an ex-math-teacher told me.
>(It was the only thing I learnt from her (sorry Els))
>
>The rules are the same, start at top, end at bottom.
>
> A
> B B
> R R R
> A A A A
> C C C C C
> A A A A A A
> D D D D D
> A A A A
> B B B
> R R
> A
>
>If you are able to solve this diamond, you will be able to solve
>any other puzzle of this type (It's just a trick).
>
>
>
> Izak van Langevelde.
** spoiler follows **
To understand the solution it may be easier to deform
the diamond a little and make it a square (for that
matter rectangle, in general).
---> A B R A C A
B R A C A D
R A C A D A
A C A D A B
C A D A B A
A D A B R A <---
So you will enter at the top left arrow and leave at the
bottom right arrow, seeking the shortest path.
There will be 5 right steps and 5 down steps. Thus possible
paths taken may be rrrbbrrbbb, brbrbbbrrr, etc. The total
possiblities is obviously
10!
--------- = 252
5! * 5!
--hhasan
In article <1990May16...@dev8c.mdcbbs.com> esc...@dev8c.mdcbbs.com writes:
>1 A 1 1 1 1 1 1 1 1 1 1 1
>2 B 2 2 2 2 2 2 2 2 2 2 1
>3 R 2 4 4 4 4 4 4 4 4 4 3
>4 A 6 8 8 8 8 8 8 8 8 7 3
>5 C 6 14 16 16 16 16 16 16 16 15 10
>6 A 20 30 32 32 32 32 32 32 31 25 10
>7 D 20 50 62 64 64 64 64 64 63 56 35
>8 A 70 112 126 128 128 128 128 127 119 91 35
>9 B 70 182 238 254 256 256 256 255 246 200 126
>10R 252 420 492 510 512 512 511 501 446 326 126
>11A 252 +672 +912 +1002+1022+1024+1023+1012+947 +772 +452 = 9090
That's no fun. How about a closed-form expression for this type of diagram?
I just couldn't resist... :-)
The last three lines should be:
70 182 238 254 256 256 256 255 246 210 126
252 420 492 510 512 512 511 501 456 336 126
252 672 912 1002 1022 1024 1023 1012 957 792 462
Total sum : 9130