Answer the multiple choice questions in the practice quiz below. If you have any trouble or feel unsure, go back over the lesson and practice any problems that might have been difficult. Be sure you understand all the concepts in this lesson before you take the actual quiz.
Good luck!
What is true of combinations?
- They cannot contain numbers.
- They can only be made up of numbers.
- Order matters.
- Order doesn't matter.
The order doesn't matter in combinations. One ordering of elements is the same as another ordering of elements.
The order doesn't matter in combinations. One ordering of elements is the same as another ordering of elements.
The order doesn't matter in combinations. One ordering of elements is the same as another ordering of elements.
The order doesn't matter in combinations. One ordering of elements is the same as another ordering of elements.
What is the equation for finding the number of combinations without repetition?
- \(\mathsf{ C(n,r) = \frac{n!}{r!(n-r)!} }\)
- \(\mathsf{ C(n,r) = \frac{(n+r+1)!}{r!(n-r)!} }\)
- \(\mathsf{ C(n,r) = \frac{(n+r-1)!}{r!(n-1)!} }\)
- \(\mathsf{ C(n,r) = \frac{r!}{n!(n-r)!} }\)
It's the same equation as the one for finding the number of permutations without repetition, except we factor out r! to account for the fact that order doesn't matter.
It's the same equation as the one for finding the number of permutations without repetition, except we factor out r! to account for the fact that order doesn't matter.
It's the same equation as the one for finding the number of permutations without repetition, except we factor out r! to account for the fact that order doesn't matter.
It's the same equation as the one for finding the number of permutations without repetition, except we factor out r! to account for the fact that order doesn't matter.
What is the equation for finding the number of combinations with repetition?
- \(\mathsf{ C(n,r) = \frac{n!}{r!(n-r)!} }\)
- \(\mathsf{ C(n,r) = \frac{(n+r+1)!}{r!(n-r)!} }\)
- \(\mathsf{ C(n,r) = \frac{(r+n-1)!}{r!(n-1)!} }\)
- \(\mathsf{ C(n,r) = \frac{r!}{n!(n-r)!} }\)
This complicated formula is derived by viewing the combination problem as a permutation of actions. There are r+n-1 actions to take, and that value is substituted into the original equation.
This complicated formula is derived by viewing the combination problem as a permutation of actions. There are r+n-1 actions to take, and that value is substituted into the original equation.
This complicated formula is derived by viewing the combination problem as a permutation of actions. There are r+n-1 actions to take, and that value is substituted into the original equation.
This complicated formula is derived by viewing the combination problem as a permutation of actions. There are r+n-1 actions to take, and that value is substituted into the original equation.
How many combinations without repetition are possible if n = 3 and r = 3?
- 27
- 9
- 2
- 1
Since repetition is not allowed, we must use the equation \(\mathsf{ \frac{n!}{r!(n-r)!} }\). So, \(\mathsf{ \frac{3!}{3!(0!)} = 1 }\).
Since repetition is not allowed, we must use the equation \(\mathsf{ \frac{n!}{r!(n-r)!} }\). So, \(\mathsf{ \frac{3!}{3!(0!)} = 1 }\).
Since repetition is not allowed, we must use the equation \(\mathsf{ \frac{n!}{r!(n-r)!} }\). So, \(\mathsf{ \frac{3!}{3!(0!)} = 1 }\).
Since repetition is not allowed, we must use the equation \(\mathsf{ \frac{n!}{r!(n-r)!} }\). So, \(\mathsf{ \frac{3!}{3!(0!)} = 1 }\).
How many combinations with repetition are allowed if n = 3 and r = 3?
- 10
- 20
- 9
- 27
Since repetition is allowed, we use the equation \(\mathsf{ \frac{(n+r-1)!}{r!(n-1)!} = \frac{5!}{3!(2!)} = 10 }\).
Since repetition is allowed, we use the equation \(\mathsf{ \frac{(n+r-1)!}{r!(n-1)!} = \frac{5!}{3!(2!)} = 10 }\).
Since repetition is allowed, we use the equation \(\mathsf{ \frac{(n+r-1)!}{r!(n-1)!} = \frac{5!}{3!(2!)} = 10 }\).
Since repetition is allowed, we use the equation \(\mathsf{ \frac{(n+r-1)!}{r!(n-1)!} = \frac{5!}{3!(2!)} = 10 }\).
Summary
Questions answered correctly:
Questions answered incorrectly:
