Combinations with Restrictions 2

Question 1 of 5

In how many ways can

2

$2$ Blue marbles and

1

$1$ Red marble be drawn from the jar below?

(30)

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Use the combinations formula to find the number of ways an item can be chosen $(r)$ $(r)$ from the total number of items $(n)$ $(n)$ .

Remember that order is not important in Combinations.

Combination Formula

_{n} C_{r} = \frac{n!}{(n - r)! r!}

$_\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!}$

First, find the ways that $2$ $2$ Blue marbles can be drawn. There are $5$ $5$ Blue marbles in the jar, which means:

r = 2

$r=2$

n = 5

$n=5$

$_\color{purple}{n}C_{\color{green}{r}}$	$=$ $=$	$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!}$	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 $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{5!}{3! 2!}$

	$=$ $=$	$\frac{5\cdot4\cdot\color{#CC0000}{3\cdot2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1}\cdot2\cdot1}$

	$=$ $=$	$\frac{20}{2}$

	$=$ $=$	$10$

Next, find the ways that $1$ $1$ Red marble can be drawn. There are $3$ $3$ Red marbles in the jar, which means:

r = 1

$r=1$

n = 3

$n=3$

$_\color{purple}{n}C_{\color{green}{r}}$	$=$ $=$	$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!}$	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 $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{3!}{2! 1!}$

	$=$ $=$	$\frac{3\cdot2!}{2!}$

	$=$ $=$	$3$ $3$	$\frac{2!}{2!}$ $(2!)/(2!)$ cancels out

Finally, multiply the two solved combinations

Number of ways

2

$2$ Blue marbles can be drawn

= 10

$=10$

Number of ways

1

$1$ Red marble can be drawn

= 3

$=3$

10 \cdot 3

$10*3$

=

$=$

30

$30$

Therefore, there are $30$ $30$ ways of drawing

2

$2$ Blue marbles and

1

$1$ Red marble from the jar.

30

$30$

Question 2 of 5

In how many ways can

1

$1$ marble of each color be drawn from the jar below?

(30)

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Use the combinations formula to find the number of ways an item can be chosen $(r)$ $(r)$ from the total number of items $(n)$ $(n)$ .

Remember that order is not important in Combinations.

Combination Formula

_{n} C_{r} = \frac{n!}{(n - r)! r!}

$_\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!}$

First, find the ways that $1$ $1$ Blue marble can be drawn. There are $5$ $5$ Blue marbles in the jar, which means:

r = 1

$r=1$

n = 5

$n=5$

$_\color{purple}{n}C_{\color{green}{r}}$	$=$ $=$	$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!}$	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 $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{5!}{4! 1!}$

	$=$ $=$	$\frac{5\cdot\color{#CC0000}{4\cdot3\cdot2\cdot1}}{\color{#CC0000}{4\cdot3\cdot2\cdot1}}$

	$=$ $=$	$5$

Next, find the ways that $1$ $1$ Red marble can be drawn. There are $3$ $3$ Red marbles in the jar, which means:

r = 1

$r=1$

n = 3

$n=3$

$_\color{purple}{n}C_{\color{green}{r}}$	$=$ $=$	$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!}$	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 $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{3!}{2! 1!}$

	$=$ $=$	$\frac{3\cdot\color{#CC0000}{2\cdot1}}{\color{#CC0000}{2\cdot1}}$

	$=$ $=$	$3$ $3$

Now, find the ways that $1$ $1$ Black marble can be drawn. There are $2$ $2$ Black marbles in the jar, which means:

r = 1

$r=1$

n = 2

$n=2$

$_\color{purple}{n}C_{\color{green}{r}}$	$=$ $=$	$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!}$	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 $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{2!}{1! 1!}$

	$=$ $=$	$2\cdot1$

	$=$ $=$	$2$ $2$

Finally, multiply the three solved combinations

Number of ways

1

$1$ Blue marble can be drawn

= 5

$=5$

Number of ways

1

$1$ Red marble can be drawn

= 3

$=3$

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

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)

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Mute

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

In how many ways can an artist buy at most

$7$ tubes of paint out of

$12$ different tubes?

(3302, 3 302, 3,302)

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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

$_\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!}$

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

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)

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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 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 combinations

Number 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.

$60$

Combinations with Restrictions 2

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