+470 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).
\(\begin{array}{ccccc} C&C&C&C&C\\ L&O&O&O&L\\ L&O&O&O&L\\ L&O&O&O&L\\ C&C&C&C&C\\ \end{array}\)
I have gotten many answers, but they are all wrong so far.
Wrong answers: 84, 64 and 4
Please help.
+23 I kind of see this as a tricky english question because if you look into the wording of the problem it's asking
How many ways can you SPELL the word cool. (not how many times can you FIND the word)
There's only one way to spell the word "cool"
So i believe the answer is 1
+633 This is a math forum, and the poster is a serious user. seriously doubt he would post a joke like that unless it was somehow indicated to be a joke thread.
+633 There are 4 that have 1 entry point from the C, and you can't use the middle C, because if you get there, there is nothing to go to: 4(1*1*3) makes 12 (you have to move diagonally away, then vertical, then you are at the middle and can go diagonally or horizontal, making 3 possibilities.)
Starting from the middle outside: 4(2 + 10 = 12) = 48.
Staring from the top and bottom center: 2(2 + 3 + 3 + 2 = 10) = 20.
I'm getting 80, but I feel like I did the last one wrong.
+633 We're all getting frustrated, so let's start with what we know:
The C's are mirrored on either side, so we know that the solution has to be even.
That's all I can think of.
+2441 I have been trying and trying and trying, and I am counting 96.
A "C" in the corner has 3 ways to make the word "cool."
Since there are 4 of them, 3*4=12
A "C" adjacent to a C in the corner has 13 ways to make the word "cool."
Since there are 4 of them, 4*13=52
A "C" in the center column has 16 ways to make the word "cool."
Since there are 2 of them, 16*2=32
Now, add these combinations together. 12+52+32 = 96 ways to make the word "cool."
Please try this answer. If this is wrong, please let me know.
+2441 I am ??!!??!?!?!?!?!!? Well, thank you!
I wish I could explain how I got that answer.
I will give my best attempt, I guess.
| Column 1 | Column 2 | Column 3 | Column 4 | Column 5 | |
| Row 1 | L | L | |||
| Row 2 | L | O | O | L | |
| Row 3 | L | O | O | O | L |
| Row 4 | C | C | C | C | C |
This is just one side of the table. I have color coded the O's such that if you land on an O as your second-to-last letter, then you will have a certain number of possibilities.
A red O means there are 3 possibilities.
A Blue O means there are 2 possibilities.
Hopefully, you can see why this is the case.
Let's start with the easiest: The "C" in the corner. Before I start, I will use my own style of Cartesian coordinates where it represents the intersection of a column and a row. For example, \((C1,R1)\) is (Column 1, Row 1), for short. In this case, this letter happens to be L.
There is only one case to consider now for the corner "C." It is the following:
1) \((C5,R4)\Rightarrow(C4,R3)\Rightarrow(C4,R2)\)
Look at that! I have landed on a red O, which signifies 3 possibilities. This means that the corner "C" has 3 possibilities. There are 4 instances of the corner "C," so 4*3=12
Time to consider the next case: The "C" adjacent to the corner "C"
Now, let's consider how many cases there are for the sequence \((C4,R4)\Rightarrow(C3,R3)\). Well, the number of possibilities is equal to the sum of the number of possibilities its neighbors are immediately adjacent to.
\((C3,R3)\) is adjacent to 2 red and 2 blue O's. Because there are 3 possibilities for a red one and 2 possibilities for a blue one, the number of possibilities is \(3+3+2+2=10\). However, this is only one sequence. Let's consider the next sequence of \((C4,R4)\Rightarrow(C4,R3)\Rightarrow(C4,R2)\). Oh look! This is a red "O," which has 3 possibilities, so let's add the number of possibilities together.
\(10+3=13\)
There are 4 of these in the diagram, so \(13*4=52\)
Now, let's consider the center "C." Well, we can use the same logic as before to know that \((C3,R4)\Rightarrow(C3,R3)\) has 10 possibilities. We know that there are 2 avenues to red O's, which equals 6 additional paths.
In total, that equates to \(10+6=16\) ways. There are two instances of there, so \(2*16=32\)
The last step is to add the numbers together. \(12+52+32=96\) ways.
