In how many ways can you spell the word COOL in the grid below? You can start on any letter, then on each step you can step one letter in any direction (up, down, left, right, or diagonal).![\[\begin{array}{ccccc} C&C&C&C&C\\ L&O&O&O&L\\ L&O&O&O&L\\ L&O&O&O&L\\ C&C&C&C&C\\ \end{array}\]](/api/ssl-img-proxy?src=http%3A%2F%2Flatex.artofproblemsolving.com%2F9%2Fb%2F6%2F9b65184c22449f7e47a02184fff4faf26e109a0b.png)
+118609 This question is very COOL. 乂⍱‿●乂
I worked out how many could be made from the top C's and then doubled the answer for the bottom C's
NOW
First I looked at the first row of Os One of those has to be the first O
the first O can come from 3 different Cs and it can lead onto 3 different OLs. So that is 9
The same story for the last O . So that is 9 too.
Now the middle O. It too can come from 3 different Cs, and it can lead onto 4 OLs on the left and another 4 on the right so that is 3(4+4)=24 outcomes.
So we have 9+24+9 = 42 from the top Cs and another 42 from the bottom Cs.
That makes 84 ways altogether.
Can I have a gold star now please? ✿乂◕‿◕乂
+118609 This question is very COOL. 乂⍱‿●乂
I worked out how many could be made from the top C's and then doubled the answer for the bottom C's
NOW
First I looked at the first row of Os One of those has to be the first O
the first O can come from 3 different Cs and it can lead onto 3 different OLs. So that is 9
The same story for the last O . So that is 9 too.
Now the middle O. It too can come from 3 different Cs, and it can lead onto 4 OLs on the left and another 4 on the right so that is 3(4+4)=24 outcomes.
So we have 9+24+9 = 42 from the top Cs and another 42 from the bottom Cs.
That makes 84 ways altogether.
Can I have a gold star now please? ✿乂◕‿◕乂
This is from AoPS.
I saw this problem.
Here is the REAL solution:
We will use casework and constructive counting. However, instead of constructing the paths C--O--O--L in that order, we will go in reverse order. This reduces our cases from 3 cases for C (corner; center; between corner and center) to 2 cases for L (center; and adjacent to center).
We consider the L in the left-center. If the first move is diagonally up right or diagonally down right, the second move must be to the right, and the last move must be to one of the 3 adjacent C's. So this subcase contains paths. If instead the first move is to the right, we have a choice of 4 O's for the next move, each of which has 3 choices for the C. This subcase has paths. Altogether, there are paths using the left-center L. By symmetry, there are also 18 paths with the right-center L.
Now we consider the L just above the left-center L. If the first move is to the right, the second move must be another move to the right, and the last move be one of the 3 adjacent C's. This subcase contains paths. If instead the first move is diagonally down right, we have 4 choices for the next O, each of which has 3 choices for the C. This subcase has paths. So there are paths using this L and, by symmetry, the other three L's that are not the left or right center.
Adding the cases, there are a total of ways of spelling COOL.
