Question

An instructor gives her class the choice to do 7 questions out of the 10 on an exam.

Answer 1

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:

$nCk=\frac{n!}{k!(n-k)!}$

So, the number of ways in which a student can select the 7 questions from the 10 questions is calculated as:

$10C7=\frac{10!}{7!(10-7)!}=120$

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:

$(5C3)(5C4)=\frac{5!}{3!(5-3)!}*\frac{5!}{4!(5-4)!}=50$

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:

$(5C4)(5C3)=\frac{5!}{4!(5-4)!}*\frac{5!}{3!(5-3)!}=50$

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:

$(5C5)(5C2)=\frac{5!}{5!(5-5)!}*\frac{5!}{2!(5-2)!}=10$

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