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Combinations with Restrictions 1Combinations 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 VideoCorrect
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 combinationsNumber 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 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 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 combinationsNumber 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 VideoCorrect
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 combinationsNumber 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?Hint
Help VideoCorrect
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 combinationsNumber 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 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
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$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 combinationsNumber 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 VideoCorrect
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 combinationsNumber 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`
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