Respuesta :
Answer:
(a) 120 choices
(b) 110 choices
Step-by-step explanation:
The number of ways in which we can select k element from a group n elements is given by:
[tex]nCk=\frac{n!}{k!(n-k)!}[/tex]
So, the number of ways in which a student can select the 7 questions from the 10 questions is calculated as:
[tex]10C7=\frac{10!}{7!(10-7)!}=120[/tex]
Then each student have 120 possible choices.
On the other hand, if a student must answer at least 3 of the first 5 questions, we have the following cases:
1. A student select 3 questions from the first 5 questions and 4 questions from the last 5 questions. It means that the number of choices is given by:
[tex](5C3)(5C4)=\frac{5!}{3!(5-3)!}*\frac{5!}{4!(5-4)!}=50[/tex]
2. A student select 4 questions from the first 5 questions and 3 questions from the last 5 questions. It means that the number of choices is given by:
[tex](5C4)(5C3)=\frac{5!}{4!(5-4)!}*\frac{5!}{3!(5-3)!}=50[/tex]
3. A student select 5 questions from the first 5 questions and 2 questions from the last 5 questions. It means that the number of choices is given by:
[tex](5C5)(5C2)=\frac{5!}{5!(5-5)!}*\frac{5!}{2!(5-2)!}=10[/tex]
So, if a student must answer at least 3 of the first 5 questions, he/she have 110 choices. It is calculated as:
50 + 50 + 10 = 110
