Coins with Combinations Me and my friend each ﬂip a coin 6 times. How many ways are there for us to end up with precisely the same number of heads? 
I know we have to use combinations but im stuck because im not sure if we have to use probabilities for heads?
 A: let $W(n)$ be the number of ways both of you get n coins. then what you want is $\sum_{k=0}^6 W(k)$
Now how many ways are there for both of them to get n heads. there are $\binom{6}{n}$ ways for you to get n heads and $\binom{6}{n}$ ways for your fiend to get n heads , therefore $W(n)=\binom{6}{n}^2$.
So what you want is $\sum_{k=0}^6\binom{6}{n}^2$
Now take a look and see that $\binom{6}{n}^2=\binom{6}{n}\cdot \binom{6}{6-n}$ Where the first term can be viewed as the number of ways to walk on a lattice(moving only up and right) to get to point (n,6-n) and the second term is the number of ways to get from that point to the point (6,6). Since every path on the lattice with moves only up and right pass through the diagonal exactly once we can see the sum of all these terms is the number of paths from 0,0 to n,n moving only up and right on the lattice.But this is also the number of ways to chose 6 moves up out of 12 moves total$\binom{12}{6}$. Therefore $\sum_{n=0}^6\binom{6}{n}^2=\binom{12}{6}=924$
A: With the given phrasing, I would say it is purely a combinatorial question:
The number of ways one of you can have $n$ heads in 6 tosses is $K_n=\binom{6}{n}$. Each of you can have either of those $K_n$ tossing patterns in order to get $n$ heads both. This makes $(K_n)^2$ kombinations for you to end having $n$ heads both. Now sum this for $n=0$ to $6$:
$$
\sum_{n=0}^6\binom{6}{n}^2=924
$$
A: Hint:  You get the same number of heads if (you get zero and he gets zero) or (you get one and he gets one) or ...  For the second, count the number of ways that you can get one and multiply by the number of ways he can get one.  Add them all up.  You were asked for the number of ways, not the probability.
