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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
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Excellent!
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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