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Combinations with Restrictions 2Combinations with Restrictions 2
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Question 1 of 5
1. Question
In how many ways can `2` Blue marbles and `1` Red marble be drawn from the jar below?
- (30)
Hint
Help VideoCorrect
Excellent!
Incorrect
Use the combinations formula to find the number of ways an item can be chosen `(r)` from the total number of items `(n)`.Remember that order is not important in Combinations.Combination Formula
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$First, find the ways that `2` Blue marbles can be drawn. There are `5` Blue marbles in the jar, which means:`r=2``n=5`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{5}C_{\color{green}{2}}$$ `=` $$\frac{\color{purple}{5}!}{(\color{purple}{5}-\color{green}{2})!\color{green}{2}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{5!}{3! 2!}$$ `=` $$ \frac{5\cdot4\cdot\color{#CC0000}{3\cdot2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1}\cdot2\cdot1}$$ `=` $$\frac{20}{2}$$ `=` $$10$$ Next, find the ways that `1` Red marble can be drawn. There are `3` Red marbles in the jar, which means:`r=1``n=3`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{3}C_{\color{green}{1}}$$ `=` $$\frac{\color{purple}{3}!}{(\color{purple}{3}-\color{green}{1})!\color{green}{1}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{3!}{2! 1!} $$ `=` $$ \frac{3\cdot2!}{2!} $$ `=` `3` `(2!)/(2!)` cancels out Finally, multiply the two solved combinationsNumber of ways `2` Blue marbles can be drawn`=10`Number of ways `1` Red marble can be drawn `=3``10*3` `=` `30` Therefore, there are `30` ways of drawing `2` Blue marbles and `1` Red marble from the jar.`30` -
Question 2 of 5
2. Question
In how many ways can `1` marble of each color be drawn from the jar below?
- (30)
Hint
Help VideoCorrect
Nice Job!
Incorrect
Use the combinations formula to find the number of ways an item can be chosen `(r)` from the total number of items `(n)`.Remember that order is not important in Combinations.Combination Formula
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$First, find the ways that `1` Blue marble can be drawn. There are `5` Blue marbles in the jar, which means:`r=1``n=5`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{5}C_{\color{green}{1}}$$ `=` $$\frac{\color{purple}{5}!}{(\color{purple}{5}-\color{green}{1})!\color{green}{1}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{5!}{4! 1!}$$ `=` $$ \frac{5\cdot\color{#CC0000}{4\cdot3\cdot2\cdot1}}{\color{#CC0000}{4\cdot3\cdot2\cdot1}}$$ `=` $$5$$ Next, find the ways that `1` Red marble can be drawn. There are `3` Red marbles in the jar, which means:`r=1``n=3`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{3}C_{\color{green}{1}}$$ `=` $$\frac{\color{purple}{3}!}{(\color{purple}{3}-\color{green}{1})!\color{green}{1}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{3!}{2! 1!} $$ `=` $$ \frac{3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}} $$ `=` `3` Now, find the ways that `1` Black marble can be drawn. There are `2` Black marbles in the jar, which means:`r=1``n=2`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{2}C_{\color{green}{1}}$$ `=` $$\frac{\color{purple}{2}!}{(\color{purple}{2}-\color{green}{1})!\color{green}{1}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{2!}{1! 1!} $$ `=` $$2\cdot1$$ `=` `2` Finally, multiply the three solved combinationsNumber of ways `1` Blue marble can be drawn`=5`Number of ways `1` Red marble can be drawn `=3`Number of ways `1` Black marble can be drawn `=2``5*3*2` `=` `30` Therefore, there are `30` ways of drawing `1` marble of each color from the jar.`30` -
Question 3 of 5
3. Question
You are asked to buy at least `6` flavors of chips out of `10` total flavors. How many combinations of chips can you buy?- (386)
Hint
Help VideoCorrect
Keep Going!
Incorrect
Use the combinations formula to find the number of ways an item can be chosen `(r)` from the total number of items `(n)`.Remember that order is not important in Combinations.Combination Formula
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$Since you are asked to buy at least `6` flavors of chips, it means you can either buy `6,7,8,9` or `10` chips out of `10` total flavors of chips. Add all combinations for each number.`r=6,7,8,9,10``n=10`$$_\color{purple}{10}C_{\color{green}{6}}+_\color{purple}{10}C_{\color{green}{7}}+_\color{purple}{10}C_{\color{green}{8}}+_\color{purple}{10}C_{\color{green}{9}}+_\color{purple}{10}C_{\color{green}{10}}$$ `=` $$210+120+45+10+1$$ Use the calculator’s combination function `=` $$386$$ There are `386` combinations possible in buying at least `6` flavors of chips out of `10` total flavors.`386` -
Question 4 of 5
4. Question
In how many ways can an artist buy at most `7` tubes of paint out of `12` different tubes?- (3302, 3 302, 3,302)
Hint
Help VideoCorrect
Excellent!
Incorrect
Use the combinations formula to find the number of ways an item can be chosen `(r)` from the total number of items `(n)`.Remember that order is not important in Combinations.Combination Formula
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$Since the artist is buying at most `7` tubes of paint, it means he can either buy `0,1,2,3,4,5,6` or `7` tubes out of `12` different tubes. Add all combinations for each number.`r=0,1,2,3,4,5,6,7``n=12`$$_\color{purple}{12}C_{\color{green}{0}}+_\color{purple}{12}C_{\color{green}{1}}+_\color{purple}{12}C_{\color{green}{2}}+_\color{purple}{12}C_{\color{green}{3}}+_\color{purple}{12}C_{\color{green}{4}}+_\color{purple}{12}C_{\color{green}{5}}+_\color{purple}{12}C_{\color{green}{6}}+_\color{purple}{12}C_{\color{green}{7}}$$ `=` $$1+12+66+220+495+792+924+792$$ Use the calculator’s combination function `=` $$3302$$ There are `3302` combinations possible in buying at most `7` tubes of paint out of `12` different tubes.`3302` -
Question 5 of 5
5. Question
Jessica wants to buy `2` cats and `3` dogs, where one of the dogs must be a German Shepherd. In how many ways can she choose these pets given that the pet shop has the following pets:
Note that only one of these dogs is a German Shepherd- (60)
Hint
Help VideoCorrect
Well Done!
Incorrect
Use the combinations formula to find the number of ways an item can be chosen `(r)` from the total number of items `(n)`.Remember that order is not important in Combinations.Combination Formula
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$First, find the ways that the `2` cats can be chosen. There are only `4` cats available, which means:`r=2``n=4`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{4}C_{\color{green}{2}}$$ `=` $$\frac{\color{purple}{4}!}{(\color{purple}{4}-\color{green}{2})!\color{green}{2}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{4!}{2! 2!}$$ `=` $$ \frac{4\cdot3\color{#CC0000}{\cdot2\cdot1}}{2\cdot1\color{#CC0000}{\cdot2\cdot1}}$$ `=` $$ \frac{12}{2}$$ `=` $$6$$ Next, find the different ways that `1` German Shepherd `(r)` can be chosen from `1` German Shepherd `(n)``r=1``n=1`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{1}C_{\color{green}{1}}$$ `=` $$\frac{\color{purple}{1}!}{(\color{purple}{1}-\color{green}{1})!\color{green}{1}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{1!}{0! 1!} $$ `=` `1` `0! =1` Now, find the different ways that the `2` other dogs `(r)` can be chosen from a total of `5` remaining dogs `(n)``r=2``n=5`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{5}C_{\color{green}{2}}$$ `=` $$\frac{\color{purple}{5}!}{(\color{purple}{5}-\color{green}{2})!\color{green}{2}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{5!}{3! 2!} $$ `=` $$ \frac{5\cdot4\color{#CC0000}{\cdot3\cdot2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1}\cdot2\cdot1} $$ `=` $$ \frac{20}{2} $$ `=` `10` Finally, multiply the three solved combinationsNumber of ways the `3` cats can be chosen`=6`Number of ways the German Shepherd can be chosen`=1`Number of ways the `2` dogs can be chosen`=10``6*1*10` `=` `60` Therefore, there are `60` ways of choosing `2` cats and `3` dogs where one of the dogs is a German Shepherd.`60`
Quizzes
- Factorial Notation
- Fundamental Counting Principle 1
- Fundamental Counting Principle 2
- Fundamental Counting Principle 3
- Combinations 1
- Combinations 2
- Combinations with Restrictions 1
- Combinations with Restrictions 2
- Combinations with Probability
- Basic Permutations 1
- Basic Permutations 2
- Basic Permutations 3
- Permutation Problems 1
- Permutation Problems 2
- Permutations with Repetitions 1
- Permutations with Repetitions 2
- Permutations with Restrictions 1
- Permutations with Restrictions 2
- Permutations with Restrictions 3
- Permutations with Restrictions 4