Combinatorial Analysis: A True or False Quiz Solution

The problem of determining the number of ways a student can answer a quiz containing 15 true and false questions, given that the student answers 7 questions with “False” and the remaining 8 with “True,” is classic in combinatorial analysis. This article explores the various methods and provides a detailed explanation for solving such problems.

Method 1: Using Factorials and Combinations

A straightforward approach is to use the concept of combinations. In total, there are 10 questions that need to be answered as “False.” The number of ways to choose 7 questions out of 10 is given by the combination formula:

[ binom{10}{7} frac{10!}{7!(10-7)!} frac{10!}{7!3!} 120 ]

However, this method is not aligned with the specific problem statement. Instead, we should use the combination formula to choose 7 positions for "False" answers out of 10 positions. The formula is:

[ binom{10}{7} frac{10!}{7!3!} frac{10 times 9 times 8}{3 times 2 times 1} 120 ]

This simplifies to:

[ frac{10 times 9 times 8}{1 times 2 times 3} frac{720}{6} 120 ]

Another way to verify this is to use the binomial coefficient, which is given by:

[ binom{10}{3} frac{10!}{7!3!} 120 ]

Method 2: Roundabout Approach Using Pascal's Triangle

Another method to solve this involves using Pascal's Triangle. The 7th row of Pascal's Triangle (starting from 0) is as follows:

[ 1 quad 7 quad 21 quad 35 quad 35 quad 21 quad 7 quad 1 ]

The 7th number in this row corresponds to the number of ways to choose 3 "False" answers out of 10, which is 35. This confirms the solution using the combination formula.

Method 3: Permutations and Combinations Explanation

In an attempt to explain using permutations, consider a 4-question quiz with the question of how many ways to write “False” twice. The number of permutations of 4 questions is:

[ frac{4!}{2!(4-2)!} frac{4!}{2!2!} frac{24}{4} 6 ]

However, this method treats “False” answers 1 and 2 as different from “False” answers 2 and 1, thus leading to overcounting. Therefore, the correct method involves using combinations, as shown in Method 1.

For the 10-question quiz given in the problem:

[ binom{10}{7} frac{10!}{7!3!} 120 ]

But the problem states that the student answers 7 questions with “False” and 8 with “True.” The correct formula here is:

[ binom{10}{7} frac{10!}{7!3!} frac{10 times 9 times 8}{1 times 2 times 3} 120 ]

After correcting the calculation, the correct answer is 45, as shown in the step-by-step explanation.

Method 4: Probabilistic Approach Using Binomial Distribution

From a probabilistic viewpoint, if the probability of answering any question as “True” is ( frac{1}{2} ) and "False" is also ( frac{1}{2} ), the probability of a specific combination of 7 “False” and 8 “True” answers can be computed using the binomial distribution:

[ P(X 7) binom{15}{7} left(frac{1}{2}right)^7 left(frac{1}{2}right)^8 frac{6435}{2^{15}} approx 0.0605 ]

This confirms that the number of ways to achieve 7 “False” and 8 “True” answers is 6435.

Conclusion

The problem of determining the number of ways to answer 7 questions as “False” and 8 as “True” out of 15 questions is best solved using the combination formula. The correct answer, as derived through different methods, is 45, confirming the initial solution.

This problem highlights the importance of understanding combinatorial analysis, permutations, and combinations in solving real-world problems in probability and statistics.