Math Problem Statement

A student must answer exactly eight questions out of ten on a final examination. (a) In how many ways can she choose the questions to answer?

In how many ways she can choose questions to answer if she must answer at least three of the last five questions?

Solution

Part (a): Choosing 8 questions out of 10

The number of ways the student can choose 8 questions out of 10 is given by the combination formula:

C(n,r)=n!r!(nr)!C(n, r) = \frac{n!}{r!(n-r)!}

Here, n=10n = 10 and r=8r = 8. Substituting these values:

C(10,8)=10!8!(108)!=10921=45C(10, 8) = \frac{10!}{8! \cdot (10-8)!} = \frac{10 \cdot 9}{2 \cdot 1} = 45

Thus, the student can choose the questions in 45 ways.


Part (b): Choosing 8 questions with at least 3 from the last 5

Here, we split the problem into cases based on how many of the last 5 questions the student answers. Let’s denote the last 5 questions as A1,A2,A3,A4,A5A_1, A_2, A_3, A_4, A_5 and the first 5 questions as B1,B2,B3,B4,B5B_1, B_2, B_3, B_4, B_5.

The student must answer at least 3 questions from the last 5 (AA), so the possible cases are:

  1. Answering 3 questions from AA and 5 questions from BB,
  2. Answering 4 questions from AA and 4 questions from BB,
  3. Answering all 5 questions from AA and 3 questions from BB.

Case 1: 3 questions from AA, 5 questions from BB

C(5,3)C(5,5)=(53)(55)=101=10C(5, 3) \cdot C(5, 5) = \binom{5}{3} \cdot \binom{5}{5} = 10 \cdot 1 = 10

Case 2: 4 questions from AA, 4 questions from BB

C(5,4)C(5,4)=(54)(54)=55=25C(5, 4) \cdot C(5, 4) = \binom{5}{4} \cdot \binom{5}{4} = 5 \cdot 5 = 25

Case 3: 5 questions from AA, 3 questions from BB

C(5,5)C(5,3)=(55)(53)=110=10C(5, 5) \cdot C(5, 3) = \binom{5}{5} \cdot \binom{5}{3} = 1 \cdot 10 = 10

Adding these cases together, the total number of ways is:

10+25+10=4510 + 25 + 10 = 45


Final Results:

  • (a) The student can choose the questions in 45 ways.
  • (b) The student can choose the questions in 45 ways if she must answer at least 3 of the last 5.

Would you like further details or explanations?


Questions for Further Exploration:

  1. How does the combination formula work, and why is it appropriate here?
  2. What are the key differences between permutations and combinations?
  3. How would the solution change if the student had to answer at least 4 of the last 5 questions?
  4. How would we approach the problem if the student could answer any number of questions from 8 to 10?
  5. What happens if the student had additional constraints, such as answering at least 2 questions from the first 5?

Tip:

When solving combination problems, break them into smaller cases if there are additional constraints. This ensures accuracy and clarity in counting possibilities.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Combinatorics
Binomial Coefficients

Formulas

Combination formula: C(n, r) = n! / (r!(n-r)!)

Theorems

Basic Counting Principles in Combinatorics

Suitable Grade Level

Grades 9-12