Counting the number of ways a test can be answered where at least one T/F and at least one multiple choice question must be answered An exam has $15$ questions: eight true/false questions and seven multiple choice questions.  You are asked to answer five of them, but your professor requires you to answer at least one true/false question and at least one multiple choice question. How many ways can you choose the questions you plan to answer.
step1. I chose $1$ question of true false type. $8$ ways of doing so.
step2. I chose $1$ question of MCQ type. $7$ ways of doing so.
step3. I chose $3$ questions of the remaining $13$ questions. $13 \cdot 12 \cdot 11$ ways of doing so.
step4. Hence, $8 \cdot 7 \cdot 13 \cdot 12 \cdot 11$ gives number of sequences of $5$ questions from a collection of $15$ questions such that there is at least one question from t/f type and mcq type each.
step5. Let the number of collections of $5$ questions from a collection of $15$ questions such that there is at least one question from t/f type and mcq type each be $N$.
step6. $N \cdot 5! = 8 \cdot 7 \cdot 11 \cdot 12 \cdot 13$
step7. $N = \dfrac{8 \cdot 7 \cdot 11 \cdot 12 \cdot 13}{5!}$
But this number above is not even a whole number.  Where did I go wrong?
 A: You overcount all of the cases.  For a case where you answer two true/false questions and three multiple choice questions, there are two way to pick which one you count in step 1 and three in step 2, so this will get counted six times.  
The easier way is to count the number of ways you can choose any five questions, then subtract the ones that are all the same kind.
A: You don't have $5!$ ways of arranging the $5$ questions.  You choose the questions specifically so that the first question was a TF and the second was MC.  So the numbers of ways to arrange your five question must be that the first is TF and the second MC.
That is, the reason you divide by $5!$ in the first place is that you need to account that picking $ABCDE$ is the same thing as picking $CEDAB$.  But unless $C$ is TF and E is MFC picking $CEDAB$ was never an option.
The problem is the number of ways or arranging the $5$ questions depends on knowing how many TF and MC questions there are.
But to choose the number of way you can have $n$ TF question and $5-n$ MC questions first.  You can have $1$ to $4$ TF questions and there are ${8 \choose n}$ ways of picking the $n$ TF question s an there are ${7\choose 5-n}$ ways of choosing the $5-n$ MC questions.
So the answer is $\sum_{n=1}^4 {8\choose n}{7\choose 5-n}=$
${8\choose 1}{7\choose 4}+{8\choose 2}{7\choose 3}+{8\choose 3}{7\choose 2}+{8\choose 4}{7\choose 1}=$
$8\frac {7!}{3!4!} + \frac {8!7!}{2!6!3!4!} + \frac {8!7!}{5!3!5!2!} + 7*\frac {8!}{4!4!}=$
$8*(5*7) + (4*7)(5*7) + (8*7)(7*3) + 7*(2*7*5)=$
$7*2(20 + 70 + 84 + 35)= 2926$
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And from Ross Millikan's answer:
"The easier way is to count the number of ways you can choose any five questions, then subtract the ones that are all the same kind."
D'oh!
#any 5 - #all 5 TF - #all 5 mc =
${15 \choose 5} - {8\choose 5} - {7\choose 5}=$
$\frac {15!}{10!5!} - \frac {8!}{5!3!} - \frac {7!}{5!2!} =$
$3003 - 56- 21 = 2926$.
