Binomial distribution ). May be this is so easy but i m not getting how to do that.
My question is..
In a party, 15 people were asked to select a number from 1 to 10, randomly. What is the probability 0f 3 out of 15 are choosing number 7. If we want to solve it using binomial then we have all the information required like n=15, p= 1/10, q= 9/10. Using this we can find P(3). What if I don't want to use binomial distribution. How can I solve this question then?
The way which I was trying is ...
Success,Failure,Number of cases 

0, 15,1

1,14,15

2,  13, 105

3,  12, 455

4,  11, 1365

5,  10, 3003

6,  9,  5005

7,  8,  6435

8,  7,  6435

9,  6,  5005

10, 5,  3003

11, 4,  1365

12, 3,  455

13, 2,  105

14, 1,  15

15, 0,  1

    total number of cases= 32768                

According to me p(3)= 455*(1/ 32768)
But this answer is totally different from the answer, using binomial formula. 
I don't know where I am wrong. Plz help me :( 
Thank you.
 A: The count of $455$ is correct. However, "success" and "failure" are not equally likely. Dividing $455$ by $2^{15}$ essentially assumes that we are tossing a fair coin. 
For example, the calculation assumes that $3$ successes and $12$ failures is just as likely as $12$ successes and $3$ failures. But it is clear that $12$ people choosing $7$ is extremely unlikely, while $3$ people choosing $7$ is not rare. 
Added Record the choices of the $15$ people as a string of length $15$, made up of the "letters" $1$ to $10$. There are $10^{15}$ such strings, and they are all equally likely. We count the number of strings that have exactly $3$ $7$'s. The location of the $7$'s can be chosen in $\binom{15}{3}$ ways. Once we have done this, we have to fill the remaining $12$ slots with non-$7$'s. This can be done in $9^{12}$ ways. So our probability is
$$\frac{\binom{15}{3}9^{12}}{10^{15}}.$$
Alternately and almost equivalently, the number of successes has Binomial Distribution. The probability of exactly $3$ successes is $\binom{15}{3}\left(\frac{1}{10}\right)^3\left(\frac{9}{10}\right)^{12}$. 
