Probability of correct answers I was wondering if anyone could shed some light on this problem that I haven't been able to solve.
$``$A test consists of 10 multiple-choice questions. The first 4 questions with 5 answers and the other 6 questions with 4 answers. In each case, only one answer is correct. Determine the probability of answering correctly exactly 3 questions.$"$
I should state beforehand that I cannot use the binomial probability formula given that according to my teacher if she hasn't taught us that, we cannot use it. We're supposed to only use multiplication rule, combination or permutation.
What I've tried so far in order to solve this problem is:


*

*We have two groups of questions. The first group (4 questions) has 5 possible answers each question. And the second group (6 questions) has 4 possible answers each question.
Therefore my sample space would be: $$(5)^4*(4)^6=2560000$$ 

*If I am to only have 3 answers then said answers could come from either the first group or the second group:



For 3 answers from the first group and zero answers from the second.$$\begin{pmatrix}
        4 \\
        3  \\
        \end{pmatrix}*\begin{pmatrix}
        6 \\
        0  \\
        \end{pmatrix}=4$$
  For 2 answers from the first group and one answer from the second.$$\begin{pmatrix}
        4 \\
        2  \\
        \end{pmatrix}*\begin{pmatrix}
        6 \\
        1  \\
        \end{pmatrix}=36$$
  For 1 answer from the first group and two answers from the second.$$\begin{pmatrix}
        4 \\
        1  \\
        \end{pmatrix}*\begin{pmatrix}
        6 \\
        2  \\
        \end{pmatrix}=60$$
  For zero answers from the first group and 3 answers from the second.$$\begin{pmatrix}
        4 \\
        0  \\
        \end{pmatrix}*\begin{pmatrix}
        6 \\
        3  \\
        \end{pmatrix}=20$$

In total I have 120 ways of getting 3 correct answers from both groups.
The possible ways I can arrange those three questions in the set of 10 questions is:
$$\begin{pmatrix}
        10 \\
        3  \\
        \end{pmatrix}=120$$
Now the probability would be $${120*120\over2560000}\approx 0.005625$$
Which I'm pretty sure it's not the right answer, but I can't seem to figure out what I'm doing wrong. Any help would be highly appreciated.
 A: You say for 3 answers from the first group and zero answers from the second there are 4 possibilities, as there are 4 answers you could get wrong in the first group. You have to multiply this by $3^6$ as there are 3 wrong answers for each of the second group questions, and there are $3^6$ the wrong answers could be done if you get none right. You forget to account for the multitude of wrong answers for each of these. That should help.
A: The problem is you're not counting correctly.  E.g. you get three correct from the first four and all of the next six wrong.  So how many ways is that?  Choose the three right answers from the four: $\binom{4}{3} = 4$.  For each of those four ways, there are four ways to get the last question wrong (because the last question has four wrong answers and thus four ways to get it wrong)!  And the other six each have three ways of getting them wrong.  So the number of ways to get exactly three right from the first four is:
$$
\binom{4}{3}\cdot 4\cdot 3^6 = 11,664
$$
For getting two right from the first group and one right from the second you have:


*

*First Group
$$
\binom{4}{2} 4^2 = 96
$$

*Second Group
$$
\binom{6}{1}3^5 = 1,458
$$


So the total number of ways would be $96\cdot1,458 = 139,968$  Then you can do the last two.  You should end up with:
$$
p \approx 23.47\%
$$
