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Rand al'Thor
Rand al'Thor
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The answer is

$3^5\times4=972$$252$ possible paths,

becausethe proof being as follows. (Thanks to @El-Guest for finding the error in my previous reasoning!)

youYou must start at the central $D$, go to $I$ in one of 4 possible ways, then from each letter you have either (A) 3 possible choices for the next one, if you're still on one of the orthogonal lines from the centre, or (B) 2 possible choices for the next one, if you've left those orthogonal lines. Also if you've left those orthogonal lines, you can't get back to them.

So, let $k$ ($1\leq k\leq6$) be the number of steps taken on those orthogonal lines. There are then just 4 possibilities for the first $k$ steps, and each of the remaining $6-k$ steps can be taken in 2 possible ways. So the total number of possibilities, for each given value of $k$, is $2^{6-k}\times4$.

Then the total is $$\sum_{k=1}^62^{6-k}\times4=4(32+16+8+4+2+1)=4\times63=252$$

This assumes that rotations and reflections of the same path count as different from each other.

For a general word of $n$ letters, laid out in a diamond configuration like this, the answer will be *at least*

$3^{n-2}\times4$,$$\sum_{k=1}^{n-1}2^{n-1-k}\times4=4(2^n-1)=2^{n+2}-4,$$

but it may be more if doubling back on the word is possible, e.g. for palindromic words or words like BANANA. In this case, the word DIAMOND can only be spelled out starting from the centre and going to the edge, which makes counting the possibilities easier.

The answer is

$3^5\times4=972$

because

you must start at the central $D$, go to $I$ in one of 4 possible ways, then from each letter you have 3 possible choices for the next one.

This assumes that rotations and reflections of the same path count as different from each other.

For a general word of $n$ letters, laid out in a diamond configuration like this, the answer will be *at least*

$3^{n-2}\times4$,

but it may be more if doubling back on the word is possible, e.g. for palindromic words or words like BANANA. In this case, the word DIAMOND can only be spelled out starting from the centre and going to the edge, which makes counting the possibilities easier.

The answer is

$252$ possible paths,

the proof being as follows. (Thanks to @El-Guest for finding the error in my previous reasoning!)

You must start at the central $D$, go to $I$ in one of 4 possible ways, then from each letter you have either (A) 3 possible choices for the next one, if you're still on one of the orthogonal lines from the centre, or (B) 2 possible choices for the next one, if you've left those orthogonal lines. Also if you've left those orthogonal lines, you can't get back to them.

So, let $k$ ($1\leq k\leq6$) be the number of steps taken on those orthogonal lines. There are then just 4 possibilities for the first $k$ steps, and each of the remaining $6-k$ steps can be taken in 2 possible ways. So the total number of possibilities, for each given value of $k$, is $2^{6-k}\times4$.

Then the total is $$\sum_{k=1}^62^{6-k}\times4=4(32+16+8+4+2+1)=4\times63=252$$

This assumes that rotations and reflections of the same path count as different from each other.

For a general word of $n$ letters, laid out in a diamond configuration like this, the answer will be *at least*

$$\sum_{k=1}^{n-1}2^{n-1-k}\times4=4(2^n-1)=2^{n+2}-4,$$

but it may be more if doubling back on the word is possible, e.g. for palindromic words or words like BANANA. In this case, the word DIAMOND can only be spelled out starting from the centre and going to the edge, which makes counting the possibilities easier.

Source Link
Rand al'Thor
Rand al'Thor
  • 117.3k
  • 28
  • 324
  • 630

The answer is

$3^5\times4=972$

because

you must start at the central $D$, go to $I$ in one of 4 possible ways, then from each letter you have 3 possible choices for the next one.

This assumes that rotations and reflections of the same path count as different from each other.

For a general word of $n$ letters, laid out in a diamond configuration like this, the answer will be *at least*

$3^{n-2}\times4$,

but it may be more if doubling back on the word is possible, e.g. for palindromic words or words like BANANA. In this case, the word DIAMOND can only be spelled out starting from the centre and going to the edge, which makes counting the possibilities easier.