Combinations with Restrictions 1

Question 1 of 6

How many ways can a soccer team of

11

$11$ players be selected from a squad of

13

$13$ existing players if the captain must be included?

(66)

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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 the captain can be selected. There is only one captain, which means:

r = 1

$r=1$

n = 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 $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{1!}{0! 1!}$
	$=$ $=$	$1$	$0! = 1$ $0! =1$

Keeping in mind that the captain has already been picked, find the different ways that the $10$ $10$ other players $(r)$ $(r)$ can be selected from a total of $12$ $12$ players $(n)$ $(n)$

r = 10

$r=10$

n = 12

$n=12$

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

$_\color{purple}{12}C_{\color{green}{10}}$	$=$ $=$	$\frac{\color{purple}{12}!}{(\color{purple}{12}-\color{green}{10})!\color{green}{10}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{12!}{2! 10!}$

	$=$ $=$	$\frac{12\cdot11\cdot{\color{#CC0000}{10}}\cdot{\color{#CC0000}{9}}\cdot{\color{#CC0000}{8}}\cdot{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{2\cdot1\cdot{\color{#CC0000}{10}}\cdot{\color{#CC0000}{9}}\cdot{\color{#CC0000}{8}}\cdot{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$

	$=$ $=$	$\frac{132}{2}$	Cancel like terms

	$=$ $=$	$66$

Finally, multiply the two solved combinations

Number of ways the captain can be selected

= 1

$=1$

Number of ways the other players can be selected

= 66

$=66$

1 \cdot 66

$1\cdot66$

=

$=$

66

$66$

Therefore, there are $66$ $66$ ways of selecting a soccer team of

11

$11$ players if the captain must be included.

66

$66$

Question 2 of 6

How many ways can we form a

5

$5$ man basketball team from a group of

9

$9$ players if the captain and the vice captain must be included?

(35)

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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 the captain and the vice captain can be selected. There is only one captain and one vice captain, which means:

r = 2

$r=2$

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

$_\color{purple}{2}C_{\color{green}{2}}$	$=$ $=$	$\frac{\color{purple}{2}!}{(\color{purple}{2}-\color{green}{2})!\color{green}{2}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{2!}{0! 2!}$
	$=$ $=$	$\frac{2\cdot1}{2\cdot1}$	$0! = 1$ $0! =1$
	$=$ $=$	$1$

Keeping in mind that the captain and the vice captain have already been picked, find the different ways that the $3$ $3$ other players $(r)$ $(r)$ can be selected from a total of $7$ $7$ players $(n)$ $(n)$

r = 3

$r=3$

n = 7

$n=7$

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

$_\color{purple}{7}C_{\color{green}{3}}$	$=$ $=$	$\frac{\color{purple}{7}!}{(\color{purple}{7}-\color{green}{3})!\color{green}{3}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{7!}{4! 3!}$

	$=$ $=$	$\frac{7\cdot6\cdot5\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}\cdot3\cdot2\cdot1}$

	$=$ $=$	$\frac{7\cdot{\color{#CC0000}{6}}\cdot5}{{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot1}$	Cancel like terms

	$=$ $=$	$\frac{7\cdot5}{1}$	$3 \times 2 = 6$ $3xx2=6$

	$=$ $=$	$35$

Finally, multiply the two solved combinations

Number of ways the captain and the vice captain can be selected

= 1

$=1$

Number of ways the other players can be selected

= 35

$=35$

1 \cdot 35

$1\cdot35$

=

$=$

35

$35$

Therefore, there are $35$ $35$ ways of selecting a basketball team of

5

$5$ players if the captain and the vice captain must be included.

35

$35$

Question 3 of 6

How many ways can students answer

6

$6$ items on a

10

$10$ items quiz if they are required to answer the first two questions?

(70)

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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 the two questions can be answered. There are only two required questions to answer, which means:

r = 2

$r=2$

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

$_\color{purple}{2}C_{\color{green}{2}}$	$=$ $=$	$\frac{\color{purple}{2}!}{(\color{purple}{2}-\color{green}{2})!\color{green}{2}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{2!}{0! 2!}$
	$=$ $=$	$\frac{2\cdot1}{2\cdot1}$	$0! = 1$ $0! =1$
	$=$ $=$	$1$

Keeping in mind that the students already answered two questions, find the different ways that the $4$ $4$ other questions $(r)$ $(r)$ can be answered from a total of $8$ $8$ remaining questions $(n)$ $(n)$

r = 4

$r=4$

n = 8

$n=8$

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

$_\color{purple}{8}C_{\color{green}{4}}$	$=$ $=$	$\frac{\color{purple}{8}!}{(\color{purple}{8}-\color{green}{4})!\color{green}{4}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{8!}{4! 4!}$

	$=$ $=$	$\frac{6\cdot7\cdot6\cdot5\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}\cdot4\cdot3\cdot2\cdot1}$

	$=$ $=$	$\frac{{\color{#CC0000}{8}}\cdot7\cdot6\cdot5}{{\color{#CC0000}{4}}\cdot3\cdot{\color{#CC0000}{2}}\cdot1}$	Cancel like terms

	$=$ $=$	$\frac{7\cdot6\cdot5}{3\cdot1}$	$4 \times 2 = 8$ $4xx2=8$

	$=$ $=$	$\frac{210}{3}$	$4 \times 2 = 8$ $4xx2=8$

	$=$ $=$	$70$

Finally, multiply the two solved combinations

Number of ways the first two questions can be answered

= 1

$=1$

Number of ways other questions can be answered

= 70

$=70$

1 \cdot 70

$1\cdot70$

=

$=$

70

$70$

Therefore, there are $70$ $70$ ways of answering

6

$6$ items on a

10

$10$ items quiz if the students are required to answer the first two questions.

70

$70$

Question 4 of 6

How many ways can a

12

$12$ -member panel be formed from a total pool of

38

$38$ people, if

2

$2$ of them are certain to be part of that panel?

1. $4232859$ $4 232 859$
2. $1004298100$ $1 004 298 100$
3. $254186856$ $254 186 856$
4. $54254015$ $54 254 015$

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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 the $2$ $2$ people who are certain to be part of the panel can be arranged. There are only $2$ $2$ positions available for them, which means:

r = 2

$r=2$

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

$_\color{purple}{2}C_{\color{green}{2}}$	$=$ $=$	$\frac{\color{purple}{2}!}{(\color{purple}{2}-\color{green}{2})!\color{green}{2}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{2!}{0! 2!}$
	$=$ $=$	$\frac{2\cdot1}{2\cdot1}$	$0! = 1$ $0! =1$
	$=$ $=$	$1$

Keeping in mind that two people are already part of the panel, find the different ways that the $10$ $10$ other members $(r)$ $(r)$ can be chosen from a total of $36$ $36$ remaining people $(n)$ $(n)$

r = 10

$r=10$

n = 36

$n=36$

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

$_\color{purple}{36}C_{\color{green}{10}}$	$=$ $=$	$\frac{\color{purple}{36}!}{(\color{purple}{36}-\color{green}{10})!\color{green}{10}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{36!}{26! 10!}$

	$=$ $=$	$254186856$ $254 186 856$	Use the calculator’s factorial function for large numbers

Finally, multiply the two solved combinations

Number of ways the two people are arranged

= 1

$=1$

Number of ways other people can be chosen

= 254186856

$=254 186 856$

1 \cdot 254186856

$1*254 186 856$

=

$=$

254186856

$254 186 856$

Therefore, there are $254186856$ $254 186 856$ ways of forming a

12

$12$ -member panel from a total pool of

38

$38$ people, if

2

$2$ of them are certain to be part of the panel.

254186856

$254 186 856$

Question 5 of 6

In how many ways can you be dealt

4

$4$ Kings and

1

$1$ other card using a standard

52

$52$ -card deck?

(48)

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Mute

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 the $4$ $4$ King cards can be dealt. There are only $4$ $4$ King cards available, which means:

r = 4

$r=4$

n = 4

$n=4$

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

$_\color{purple}{4}C_{\color{green}{4}}$	$=$ $=$	$\frac{\color{purple}{4}!}{(\color{purple}{4}-\color{green}{4})!\color{green}{4}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{4!}{0! 4!}$

	$=$ $=$	$\frac{4\cdot3\cdot2\cdot1}{4\cdot3\cdot2\cdot1}$	$0! = 1$ $0! =1$

	$=$ $=$	$1$

Keeping in mind that

4

$4$ cards have already been dealt, find the different ways that $1$ $1$ other card $(r)$ $(r)$ can be chosen from a total of $48$ $48$ remaining cards $(n)$ $(n)$

r = 1

$r=1$

n = 48

$n=48$

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

$_\color{purple}{48}C_{\color{green}{1}}$	$=$ $=$	$\frac{\color{purple}{48}!}{(\color{purple}{48}-\color{green}{1})!\color{green}{1}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{48!}{47! 1!}$

	$=$ $=$	$\frac{48\cdot47!}{47!}$

	$=$ $=$	$48$ $48$	$\frac{47!}{47!}$ $(47!)/(47!)$ cancels out

Finally, multiply the two solved combinations

Number of ways the

4

$4$ King cards can be dealt

= 1

$=1$

Number of ways one other card can be dealt

= 48

$=48$

1 \cdot 48

$1*48$

=

$=$

48

$48$

Therefore, there are $48$ $48$ ways of dealing

4

$4$ Kings and one other card from a standard deck of

52

$52$ cards.

48

$48$

Question 6 of 6

In how many ways can you be dealt

3

$3$ Queens and

2

$2$ other cards using a standard

52

$52$ -card deck?

(4512)

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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 the $3$ $3$ Queens can be dealt. There are $4$ $4$ Queens available, which means:

r = 3

$r=3$

n = 4

$n=4$

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

$_\color{purple}{4}C_{\color{green}{3}}$	$=$ $=$	$\frac{\color{purple}{4}!}{(\color{purple}{4}-\color{green}{3})!\color{green}{3}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{4!}{1! 3!}$

	$=$ $=$	$\frac{4\color{#CC0000}{\cdot3\cdot2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1}}$	$0! = 1$ $0! =1$

	$=$ $=$	$4$

Keeping in mind that the

4

$4$ Queens cannot be chosen again, find the different ways that $2$ $2$ other cards $(r)$ $(r)$ can be chosen from a total of $48$ $48$ remaining cards $(n)$ $(n)$

r = 2

$r=2$

n = 48

$n=48$

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

$_\color{purple}{48}C_{\color{green}{2}}$	$=$ $=$	$\frac{\color{purple}{48}!}{(\color{purple}{48}-\color{green}{2})!\color{green}{2}!}$	Substitute the values of $r$ $r$ and $n$ $n$

	$=$ $=$	$\frac{48!}{46! 2!}$

	$=$ $=$	$\frac{48\cdot47\cdot46!}{46!\cdot2\cdot1}$

	$=$ $=$	$\frac{2256}{2}$ $2256/2$	$\frac{46!}{46!}$ $(46!)/(46!)$ cancels out

	$=$ $=$	$1128$ $1128$

Finally, multiply the two solved combinations

Number of ways the

3

$3$ Queens can be dealt

= 4

$=4$

Number of ways two other cards can be dealt

= 1128

$=1128$

4 \cdot 1128

$4*1128$

=

$=$

4512

$4512$

Therefore, there are $4512$ $4512$ ways of dealing

3

$3$ Queens and two other cards from a standard deck of

52

$52$ cards.

4512

$4512$

Combinations with Restrictions 1

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Quiz summary

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1. Question

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Fantastic!

Combination Formula

2. Question

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Nice Job!

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3. Question

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Great Work!

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4. Question

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Correct!

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5. Question

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Keep Going!

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6. Question

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Fantastic!

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