A student

A student is to answer 8 out of 10 questions on the exam.
a) find the number n of ways the student can choose 8 out of 10 questions
b) find n if the student must answer the first three questions
c) How many if he must answer at least 4 of the first five questions?

Final Answer:

a =  45
b =  21
c =  35

Step-by-step explanation:

$$\begin{gathered} C_8(10)=\left(\genfrac{}{}{0ex}{}{10}{8}\right)=\frac{10!}{8!(10-8)!}=\frac{10\cdot 9}{2\cdot 1}=45 \\ a=\left(\genfrac{}{}{0ex}{}{8}{10}\right)=45 \end{gathered}$$

$$\begin{gathered} C_5(7)=\left(\genfrac{}{}{0ex}{}{7}{5}\right)=\frac{7!}{5!(7-5)!}=\frac{7\cdot 6}{2\cdot 1}=21 \\ b=\left(\genfrac{}{}{0ex}{}{8-3}{10-3}\right)=21 \end{gathered}$$

$$\begin{gathered} C_4(5)=\left(\genfrac{}{}{0ex}{}{5}{4}\right)=\frac{5!}{4!(5-4)!}=\frac{5}{1}=5 \\ {C}_5(5)=\left(\genfrac{}{}{0ex}{}{5}{5}\right)=\frac{5!}{5!(5-5)!}=\frac{1}{1}=1 \\ {C}_3(5)=\left(\genfrac{}{}{0ex}{}{5}{3}\right)=\frac{5!}{3!(5-3)!}=\frac{5\cdot 4}{2\cdot 1}=10 \\ c=\left(\genfrac{}{}{0ex}{}{4}{5}\right)\cdot \left(\genfrac{}{}{0ex}{}{4}{5}\right)+\left(\genfrac{}{}{0ex}{}{5}{5}\right)\cdot \left(\genfrac{}{}{0ex}{}{5}{3}\right)=\left(\genfrac{}{}{0ex}{}{4}{5}\right)\cdot \left(\genfrac{}{}{0ex}{}{4}{5}\right)+1\cdot 10=35 \end{gathered}$$

Try another example