Combinations with Restrictions 1

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Question 1 of 6

1. Question

How many ways can a soccer team of `11` players be selected from a squad of `13` existing players if the captain must be included?

(66)

Hint

Help Video

Correct

Fantastic!

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

`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`

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

`r=10`

`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` and `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`

Number of ways the other players can be selected`=66`

$$1\cdot66$$

`=`

$$66$$

Therefore, there are `66` ways of selecting a soccer team of `11` players if the captain must be included.

`66`

Question 2 of 6

2. Question

How many ways can we form a `5` man basketball team from a group of `9` players if the captain and the vice captain must be included?

(35)

Hint

Help Video

Correct

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

`r=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` and `n`

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

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

`r=3`

`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` and `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

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

	`=`	$$35$$

Finally, multiply the two solved combinations

Number of ways the captain and the vice captain can be selected`=1`

Number of ways the other players can be selected`=35`

$$1\cdot35$$

`=`

$$35$$

Therefore, there are `35` ways of selecting a basketball team of `5` players if the captain and the vice captain must be included.

`35`

Question 3 of 6

3. Question

How many ways can students answer `6` items on a `10` items quiz if they are required to answer the first two questions?

(70)

Hint

Help Video

Correct

Great Work!

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

`r=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` and `n`

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

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

`r=4`

`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` and `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

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

	`=`	$$ \frac{210}{3} $$	`4xx2=8`

	`=`	$$70$$

Finally, multiply the two solved combinations

Number of ways the first two questions can be answered`=1`

Number of ways other questions can be answered`=70`

$$1\cdot70$$

`=`

$$70$$

Therefore, there are `70` ways of answering `6` items on a `10` items quiz if the students are required to answer the first two questions.

`70`

Question 4 of 6

4. Question

How many ways can a `12`-member panel be formed from a total pool of `38` people, if `2` of them are certain to be part of that panel?

`254 186 856`
`4 232 859`
`1 004 298 100`
`54 254 015`

Hint

Help Video

Correct

Correct!

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

`r=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` and `n`

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

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

`r=10`

`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` and `n`

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

	`=`	`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`

Number of ways other people can be chosen`=254 186 856`

`1*254 186 856`

`=`

`254 186 856`

Therefore, there are `254 186 856` ways of forming a `12`-member panel from a total pool of `38` people, if `2` of them are certain to be part of the panel.

`254 186 856`

Question 5 of 6

5. Question

In how many ways can you be dealt `4` Kings and `1` other card using a standard `52`-card deck?

(48)

Hint

Help Video

Correct

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

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

First, find the ways that the `4` King cards can be dealt. There are only `4` King cards available, which means:

`r=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` and `n`

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

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

	`=`	$$1$$

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

`r=1`

`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` and `n`

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

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

	`=`	`48`	`(47!)/(47!)` cancels out

Finally, multiply the two solved combinations

Number of ways the `4` King cards can be dealt`=1`

Number of ways one other card can be dealt`=48`

`1*48`

`=`

`48`

Therefore, there are `48` ways of dealing `4` Kings and one other card from a standard deck of `52` cards.

`48`

Question 6 of 6

6. Question

In how many ways can you be dealt `3` Queens and `2` other cards using a standard `52`-card deck?

(4512)

Hint

Help Video

Correct

Fantastic!

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

`r=3`

`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` and `n`

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

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

	`=`	$$4$$

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

`r=2`

`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` and `n`

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

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

	`=`	`2256/2`	`(46!)/(46!)` cancels out

	`=`	`1128`

Finally, multiply the two solved combinations

Number of ways the `3` Queens can be dealt`=4`

Number of ways two other cards can be dealt`=1128`

`4*1128`

`=`

`4512`

Therefore, there are `4512` ways of dealing `3` Queens and two other cards from a standard deck of `52` cards.

`4512`

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Basic Permutations 2
Basic Permutations 3
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