Five friends meeting at the same coffeehouse 
Five friends agree to meet at Joe's Coffeehouse in Capital City. However, there are $5$ different Joe's Coffeehouse locations in Capital City, and the friends neglected to agree on which one to meet at, so they each choose one at random. What is the probability that


(a) they all end up at the same Joe's?


(b) they all end up at different Joe's?

For each of those parts I solved it in two different ways and got different answers, I'm not sure which one is correct in both cases. Any explanation as to which is correct, which is wrong, and why would be very well appreciated.
(a) APPROACH 1: Let's fix a friend, there's a ${1\over5}$ chance of him going to each particular Joe's, and then for each of those $5$ options for the remaining $4$ friends, there's a ${1\over5}$ chance of going to that same Joe's, so the probability they all end up at the same Joe's is:$$\left({1\over5}\right)\left({1\over5}\right)^4 + \left({1\over5}\right)\left({1\over5}\right)^4 + \left({1\over5}\right)\left({1\over5}\right)^4 + \left({1\over5}\right)\left({1\over5}\right)^4 + \left({1\over5}\right)\left({1\over5}\right)^4 = \left({1\over5}\right)^4 = {1\over{625}}$$
APPROACH 2: The probability is going to be$${{\# \text{ of nonnegative integer solutions to }x_1 + x_2 + x_3 + x_4 + x_5 = 5 \text{ such that }x_i = 5 \text{ for some }i \text{ between }1\text{ and }5 \text{ inclusive}}\over{\# \text{ of nonnegative integer solutions to }x_1 + x_2 + x_3 + x_4 + x_5 = 5}} = {5\over{\binom{9}{5}}} = {5\over{126}}$$
(b) APPROACH 1: Let's fix a friend, there's a ${1\over5}$ chance of him going to each particular Joe's, and so for that particular Joe's, there's a ${4\over5}$ chance of the next guy going to a different Joe's, a ${3\over5}$ for the next guy etc. so the probability they all end up at different Joe's is:$$5 \left({1\over5}\right)\left({4\over5}\right)\left({3\over5}\right)\left({2\over5}\right) \left({1\over5}\right) = {{24}\over{625}}$$
APPROACH 2: The probability is going to be$${{\# \text{ of nonnegative integer solutions to }x_1 + x_2 + x_3 + x_4 + x_5 = 5 \text{ such that }x_i = 1 \text{ for all }i \text{ between }1\text{ and }5 \text{ inclusive}}\over{\# \text{ of nonnegative integer solutions to }x_1 + x_2 + x_3 + x_4 + x_5 = 5}} = {1\over{\binom{9}{5}}} = {1\over{126}}$$
 A: I am unfamiliar with your methods, but I have worked out a solution to both, two different ways, with answers that agree with your answers in approach one.
Method one:
Take the first case:
One friend will definitely go to one shop. The next friend has a $1/5$ probability of visiting that shop, as do the rest of the friends. The probability here is $(1/5)^4=1/625$.
Second case:
One friend will definitely go to one shop. The next friend has a $4/5$ probability of not visiting their shop. The next has $3/5$,... and so on; the total probability is $4!/5^4=24/625$.
Method two:
There are $5\times5\times5\dots=5^5$ total outcomes. Label shops 1-5 as letters A-E. E.g., such an outcome might be “AABED” which says which friend went to which shop.
First case:
Now the event where they are all in one shop happens five times: “AAAAA,BBBBB,...” and so we have probability $5/(5^5)=1/5^4=1/625$, which is what I calculated before.
Second case:
To be all in different shops, we have a string “ABCDE” and all its $5!$ permutations. $5!/5^5=(5\cdot 4!)/5^5=4!/5^4=24/625$ as I said before.
Hope this helps.
A: The $\binom{9}{5}$ outcomes you're considering in the second approaches aren't equally likely, so you cannot compute the probability as the number of favorable outcomes divided by the number of total outcomes.
As an analogous problem, if you flip a coin twice, you get three outcomes: 2 heads, 2 tails, or one of each; however, these categories of outcomes aren't equally likely.
A: Your method $1$ for the cases is absolutely correct, along with explanation.
Regarding method $2$, we cannot apply stars and bar method in this case. Take the example, we have to put $n$ indistinguishable balls in $k$ distinguishable cans. Here you can apply the stars and bar method because balls are indistinguishable, it doesn't matter which ball is going to which can, it does not increase the number of cases. But in the given question, all $5$ people are distinguishable. It does matter which person is going to which restaurant and it will also increase the number of cases, as described in your approach $1$.
If you still have doubt, you can go through https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) .
