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 5 questions?

Correct result:

a =  45
b =  21
c =  35

Solution:

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

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

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

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