2. Re: shortest way

why should we just solve these for you?

Show some work and ask specific questions regarding where you are stuck.

3. Re: shortest way

Originally Posted by romsek
why should we just solve these for you?
Show some work and ask specific questions regarding where you are stuck.
Because, lazy, lazy, too lazy to even type out the question to make it readable.

4. Re: shortest way

Your method is correct for the complete 4x8 rectangle. That's not the given shape. Every possible path goes through either the top-left or the bottom-right corner of the corner square.

5. Re: shortest way

yes if I can find the lost part and subtract itfrom 4X8 I will solve it but I don't know how to.

6. Re: shortest way

1 way to get from A to B, through C
A to D :4!/3!=4
8 ways from D to B =8!/7!
8x4+1=33
What do you think?

7. Re: shortest way

Originally Posted by kastamonu
1 way to get from A to B, through C
A to D :4!/3!=4
8 ways from D to B =8!/7!
8x4+1=33
What do you think?
Look at my relabeling of your diagram. A path (I) could be $ADB$, that is $A\to D\to B$ from A thru D to B.
Then path (II) is $AED$. On path (I) there is only one possibility $VVVVVHHHHHHHH$, go vertical five times to $D$ then go horizontal eight times to $B$. Doing that is a path $A\to B$. Note that means $C$ is on path (I) but $E$ is not.

Now consider path (II), $AEB$ or $A\to E\to B$. There at two parts $A\to E$ & $E\to B$.
The first part consists of $\bf{HVVV}$ in some order.
There are four ways to order that string.
$C$ may or may not be on that part, but $D$ is not.
The second part of path (II) consists of $\bf{VHHHHHHH}$ in some order There are eight ways to order that string.
$F\text{ or }G$ may or may not be on that part, but not both.

The total is $1+4\cdot 8$. Now you explain to others why!

8. Re: shortest way

Originally Posted by kastamonu
1 way to get from A to B, through C
A to D :4!/3!=4
8 ways from D to B =8!/7!
8x4+1=33
What do you think?