Permutation by grid

[ShanQ]

Original Question:
Every morning, Milly walks from her home H(0, 0) to the gym G(6, 6) along city streets that are laid out in a square grid as shown. She always takes a path of shortest distance. On her way to the gym, she often purchases a coffee at a café located at point C(2, 2). A new café opens up at point B(4, 4). How many paths can Milly take, assuming that she buys coffee at either café?

My workings:
From H (0,0) to café (2,2), then to G (6,6), # of ways X= [math]\frac{4!}{2!2!}\times \frac {8!}{4!4!} = 420[/math]From H (0,0) to café (4,4), then to G (6,0), # of ways Y = 420 [same as above]
The question asks how many paths when she buys coffee at either café.
I would take [imath]X \cup Y = X + Y - X \cap Y = 420+420- X\cap Y[/imath]
I don't know how to work out the intersection between X and Y.

FYI: the final answer is 624.

Thank you so much for your help.

 

[Dr.Peterson]

I don't know how to work out the intersection between X and Y.

How many ways are there to go from (0,0) to (2,2) to (4,4) to (6,6)?

 

[pka]

Original Question:
Every morning, Milly walks from her home H(0, 0) to the gym G(6, 6) along city streets that are laid out in a square grid as shown. She always takes a path of shortest distance. On her way to the gym, she often purchases a coffee at a café located at point C(2, 2). A new café opens up at point B(4, 4). How many paths can Milly take, assuming that she buys coffee at either café?
View attachment 33935
My workings:
From H (0,0) to café (2,2), then to G (6,6), # of ways X= [math]\frac{4!}{2!2!}\times \frac {8!}{4!4!} = 420[/math]From H (0,0) to café (4,4), then to G (6,0), # of ways Y = 420 [same as above]
The question asks how many paths when she buys coffee at either café.
I would take [imath]X \cup Y = X + Y - X \cap Y = 420+420- X\cap Y[/imath]
I don't know how to work out the intersection between X and Y.

What is [imath] 2\cdot\dfrac{4!}{(2)^2}\dfrac{8!}{(4!)^2}-\left(\dfrac{4!}{(2)^2}\right)^3=~?[/imath]

 

[Steven G]

Are you counting any path multiple time? Subtraction can take car of that.
What do you think the definition of X∩Y is?

 

[ShanQ]

Thank you pka. Could you please provide some explanation in details? Thank you.

 

[Dr.Peterson]

Thank you pka. Could you please provide some explanation in details? Thank you.

I believe his intention was to give YOU something to think about. Have you done so?

Do you notice that the first term he wrote is what you have done so far? What might the second term mean? When you put together all the help you've been given, the explanation is right in front of your eyes; we're all waiting for you to take the next step.

You did, at some point, read this, right?

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[pka]

Thank you pka. Could you please provide some explanation in details? Thank you.

Please note that Milly does buy coffee at (2,2) or (4,4) but not both(either/or)
Can you explain to us why the number paths [imath](0,0)\to(2,2)\to(6,6)[/imath]
is equal the number paths [imath](0,0)\to(4,4)\to(6,6)~?[/imath].
Is the number paths [imath](0,0)\to(2,2)~\&~(4,4)\to(6,6)\text{equal}~?[/imath].
Now, why subtract off the cube of the number paths [imath](0,0)\to(2,2)~?[/imath]
Please read the guide lines:

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Welcome to FreeMathHelp.com! Please take the time to read the following before you make your first post. It will help you to get your math questions answered promptly and in the most helpful manner. A summary is available here, but please read the complete guidelines below at your earliest...

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If you post an attempt to answer the questions I posted, then we can proceed.
imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[ShanQ]

From (0,0) →(2,2), # of ways = [math]\frac{4!}{2!\times2!}[/math]From (2,2)→(6,6), # of ways = [math]\frac{8!}{4!\times 4!}[/math]Therefore, from (0,0) to (6,6) via (2,2) X= [math]\frac{4!}{2!\times2!} \times \frac{8!}{4!\times4!}= 420[/math]
From (0,0) →(4,4), # of ways = [math]\frac{8!}{4!\times 4!}[/math]From (4,4) →(6,6), # of ways = [math]\frac{4!}{2!\times2!}[/math]Therefore, from (0,0) to (6,6) via (4,4) Y= [math]\frac{4!}{2!\times2!} \times \frac{8!}{4!\times4!}= 420[/math]
The above calculation proves that X=Y.

The question asks how many paths when she buys coffee at either café.
This means:
[math]X \cup Y = |X| + |Y| - X \cap Y = 420+420-X \cap Y[/math]
I am not sure [math]X \cap Y[/math] has covered (0,0) →(2,2), (2,2) →(4,4) and (4,4) →(6,6).
These three sections, # of ways = [math]\left (\frac{4!}{2! \times 2!} \right)^3= 216[/math]
If so, it will be
[math]X \cup Y = |X| + |Y| - X \cap Y = 840-216=624[/math]