Distributing 10 different folders between 7 people such that John gets at least 1 folder I did this by choosing 1 folder to give John ${10 \choose 1}=10$ and then saying that each of the remaining 9 folders can have 7 "states": They have to go to one of the 7 people. So that's $7^9$
So my answer is $10 \times 7^9$
My friend, however, says that the answer should be counting all the possibilites and substracting the ones in which John doesn't get any folders. That'd be $7^{10} - 6^{10}$
So who is wrong here and why? Thanks in advance
 A: Your friend is correct.  You are counting cases in which John receives more than one folder multiple times.  
If John receives exactly $k$ folders, then there are $6^{10 - k}$ ways to distribute the remaining $10 - k$ folders.  Since John can receive exactly $k$ folders in $\binom{10}{k}$ ways, the number of ways to distribute the folders so that John receives at least one is 
$$\sum_{k = 1}^{10} \binom{10}{k}6^{10 - k} = \binom{10}{1}6^9 + \binom{10}{2}6^8 + \binom{10}{3}6^7 + \binom{10}{4}6^6 + \binom{10}{5}6^5 + \binom{10}{6}6^4 + \binom{10}{7}6^3 + \binom{10}{8}6^2 + \binom{10}{9}6^1 + \binom{10}{10}6^0 = 7^{10} - 6^{10}$$ 
Say each folder is a different color and John receives blue, red, and green folders.  You count this distribution three times, once for each way of designating one of the colors as the one John receives and the other two as additional folders.  In general, if John receives $k$ folders, you count this case $k$ times, once for each way of designating a particular folder as the one reserved for John.  Notice that
$$\sum_{k = 1}^{10} k\binom{10}{k}6^{10 - k} = 1\binom{10}{1}6^9 + 2\binom{10}{2}6^8 + 3\binom{10}{3}6^7 + 4\binom{10}{4}6^6 + 5\binom{10}{5}6^5 + 6\binom{10}{6}6^4 + 7\binom{10}{7}6^3 + 8\binom{10}{8}6^2 + 9\binom{10}{9}6^1 + 10\binom{10}{10}6^0 = 10 \cdot 7^9$$ 
