Seven coins are flipped. Find the probability that three land on one side, four on the other. I'm a little confused as to which answer is correct or if either are correct.
Answer #1: $\frac{\binom{7}{3}}{2^{7}}$ 7 choose 3= number of ways three land on one side, $2^{7}$= outcomes of coin toss
Answer #2: $\frac{\binom{7}{3}+\binom{4}{4}}{2^{7}}$ 7 choose 3= number of ways three coins land on one side,4 choose 4= number of ways four coins land on the other, and  $2^{7}$= outcomes of coin toss
Can someone please double check my results? Thank you!
 A: I do believe that your first answer is nearly correct, but off by a factor of 2.
Indeed, there are 3 out of seven ways that three coins will be on a fixed side (say head) and the four other coins on the other (say tail). But you also need to count that if three coins are tail and four heads, the condition of the question is still met, so the result should be:
$$
\binom{7}{3}\cdot \frac{1}{2^6}
$$
A: Another point of view that may help you understand this problem, would be to consider the binomial distribution (for more information go to wikipedia). 
Now, let's say that, when the coin lands on one of the sides (of course it doesn't matter which one) it is a success. Therefore, since the results of each coin toss verify the binomial distribution, we have that;
$$
P(k;,n,p)=\binom{n}{k}\cdot p^k\cdot (1-p)^{n-k}\Longrightarrow P\left(3; 7, \frac{1}{2}\right)= \binom{7}{3}\cdot \frac{1}{2^7}
$$
A: I just tried solving this in a fun, silly way with my text editor:
I used multi-selection editing, a "double lines" hotkey, and copy-and-and paste to build all of the possibilities,
then I used line sorting with elastic tabstops and regex selection/find-and-replace to group them into functionally identical results.
The answer I got was 70/128
or 35/64 reduced
or 0.546875 decimal.
Is that right? xD
(
Or possibly half that if you think the phrasing is ambiguous;
ie, only counting oooxxxx,
rather than both oooxxxx and ooooxxx.
I think the latter is the more obvious interpretation.
)
All possibilites, as built:
xxxxxxx
oxxxxxx
xoxxxxx
ooxxxxx
xxoxxxx
oxoxxxx
xooxxxx
oooxxxx
xxxoxxx
oxxoxxx
xoxoxxx
ooxoxxx
xxooxxx
oxooxxx
xoooxxx
ooooxxx
xxxxoxx
oxxxoxx
xoxxoxx
ooxxoxx
xxoxoxx
oxoxoxx
xooxoxx
oooxoxx
xxxooxx
oxxooxx
xoxooxx
ooxooxx
xxoooxx
oxoooxx
xooooxx
oooooxx
xxxxxox
oxxxxox
xoxxxox
ooxxxox
xxoxxox
oxoxxox
xooxxox
oooxxox
xxxoxox
oxxoxox
xoxoxox
ooxoxox
xxooxox
oxooxox
xoooxox
ooooxox
xxxxoox
oxxxoox
xoxxoox
ooxxoox
xxoxoox
oxoxoox
xooxoox
oooxoox
xxxooox
oxxooox
xoxooox
ooxooox
xxoooox
oxoooox
xooooox
oooooox
xxxxxxo
oxxxxxo
xoxxxxo
ooxxxxo
xxoxxxo
oxoxxxo
xooxxxo
oooxxxo
xxxoxxo
oxxoxxo
xoxoxxo
ooxoxxo
xxooxxo
oxooxxo
xoooxxo
ooooxxo
xxxxoxo
oxxxoxo
xoxxoxo
ooxxoxo
xxoxoxo
oxoxoxo
xooxoxo
oooxoxo
xxxooxo
oxxooxo
xoxooxo
ooxooxo
xxoooxo
oxoooxo
xooooxo
oooooxo
xxxxxoo
oxxxxoo
xoxxxoo
ooxxxoo
xxoxxoo
oxoxxoo
xooxxoo
oooxxoo
xxxoxoo
oxxoxoo
xoxoxoo
ooxoxoo
xxooxoo
oxooxoo
xoooxoo
ooooxoo
xxxxooo
oxxxooo
xoxxooo
ooxxooo
xxoxooo
oxoxooo
xooxooo
oooxooo
xxxoooo
oxxoooo
xoxoooo
ooxoooo
xxooooo
oxooooo
xoooooo
ooooooo

Sorted:
        xxxxxxx #no
o       xxxxxx  #no
o       xxxxxx  #no
o       xxxxxx  #no
o       xxxxxx  #no
o       xxxxxx  #no
o       xxxxxx  #no
o       xxxxxx  #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
oo      xxxxx   #no
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
ooo     xxxx    #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
oooo    xxx     #yes
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
ooooo   xx      #no
oooooo  x       #no
oooooo  x       #no
oooooo  x       #no
oooooo  x       #no
oooooo  x       #no
oooooo  x       #no
oooooo  x       #no
ooooooo         #no

A: If we interpret the question to mean that three land on a specific side and four land on the other, then your first answer is correct.  To correct your second answer (under the same interpretation), note that for each of the $\binom{7}{3}$ ways for a coin to land on one side, there are $\binom{4}{4}$ ways for it to land on the other.  Hence, the numerator should be
$$\binom{7}{3}\binom{4}{4} = \binom{7}{3}$$
Edit:  After reading Nougat Rillettes' answer, I realized that we should consider the following interpretation of the problem.
If we instead interpret the question to mean that three coins land on one side of the coin and four coins land on the opposite side (without caring which side appears three times), then the probability would be 
$$\frac{2\binom{7}{3}}{2^7}$$
since there are two possible faces that could occur exactly three times.
