Questions: Question 13 Solve the problem. A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee must answer 6 of the multiple-choice questions and 4 of the open-ended problems, in how many ways can the questions and problems be chosen? 1260 261,273,600 1296 21,772,800 Question 14 Evaluate the factorial expression. 800!/799!

Solution Steps

To solve the problem of choosing questions from the exam, we need to use combinations. For the multiple-choice questions, we choose 6 out of 9, and for the open-ended problems, we choose 4 out of 6. The total number of ways to choose the questions is the product of these two combinations.

For the factorial expression, we simplify \( \frac{800!}{799!} \) by canceling out the common terms, which results in 800.

To determine the number of ways to choose 6 out of 9 multiple-choice questions, we use the combination formula:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

For our case, \( n = 9 \) and \( r = 6 \):

\[ \binom{9}{6} = \frac{9!}{6! \cdot (9-6)!} = \frac{9!}{6! \cdot 3!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]

Next, we calculate the number of ways to choose 4 out of 6 open-ended problems using the same combination formula:

\[ \binom{6}{4} = \frac{6!}{4! \cdot (6-4)!} = \frac{6!}{4! \cdot 2!} = \frac{6 \times 5}{2 \times 1} = 15 \]

The total number of ways to choose the questions is the product of the two combinations calculated:

\[ \text{Total Ways} = \binom{9}{6} \times \binom{6}{4} = 84 \times 15 = 1260 \]

For the factorial expression \( \frac{800!}{799!} \), we simplify it as follows:

\[ \frac{800!}{799!} = 800 \]

Final Answer