Combinations 1
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Question 1 of 6
1. Question
Evaluate$$_4 C_3$$- (4)
Hint
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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
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$Substitute the values to the combination formula.`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` `=` $$ \frac{4!}{1! 3!} $$ `=` $$ \frac{4\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{1\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$$ `=` $$ \frac{4}{1} $$ Cancel like terms `=` $$4$$ `4` -
Question 2 of 6
2. Question
Evaluate$$_{10} C_4$$- (210)
Hint
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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
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$Substitute the values to the combination formula.`r=4``n=10`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{4}C_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{10}!}{(\color{purple}{10}-\color{green}{4})!\color{green}{4}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{10!}{6! 4!} $$ `=` $$ \frac{10\cdot9\cdot8\cdot7\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}\cdot4\cdot3\cdot2\cdot1}$$ `=` $$ \frac{10\cdot9\cdot{\color{#CC0000}{8}}\cdot7}{{\color{#CC0000}{4}}\cdot3\cdot{\color{#CC0000}{2}}\cdot1}$$ Cancel like terms `=` $$ \frac{10\cdot9\cdot7}{3\cdot1} $$ `4xx2=8` `=` $$ \frac{630}{3} $$ `=` $$210$$ `210` -
Question 3 of 6
3. Question
Evaluate$$_9 C_2$$- (36)
Hint
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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
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$Substitute the values to the combination formula.`r=2``n=9`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{4}C_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{9}!}{(\color{purple}{9}-\color{green}{2})!\color{green}{2}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{9!}{7! 2!} $$ `=` $$ \frac{9\cdot8\cdot{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}\cdot2\cdot1}$$ `=` $$ \frac{9\cdot8}{2\cdot1} $$ Cancel like terms `=` $$ \frac{72}{2} $$ `=` $$36$$ `36` -
Question 4 of 6
4. Question
How many ways can we pick `5` paintings from a gallery with `14` paintings?- (2002)
Hint
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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
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$We need to find the different ways that `5` paintings `(r)` can be selected from a total of `14` paintings `(n)``r=5``n=14`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{13}C_{\color{green}{11}}$$ `=` $$\frac{\color{purple}{14}!}{(\color{purple}{14}-\color{green}{5})!\color{green}{5}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{14!}{9! 5!} $$ `=` $$ \frac{14\cdot13\cdot12\cdot11\cdot10\cdot{\color{#CC0000}{9}}\cdot{\color{#CC0000}{8}}\cdot{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{9}}\cdot{\color{#CC0000}{8}}\cdot{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}\cdot5\cdot4\cdot3\cdot2\cdot1} $$ `=` $$ \frac{240\;240}{120} $$ Cancel like terms `=` $$2002$$ `2002` -
Question 5 of 6
5. Question
How many ways can a soccer team of `11` players be selected from a squad of `13` existing players?
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
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$We need to find the different ways that a team of `11` players `(r)` can be selected from a total of `13` players `(n)``r=11``n=13`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{13}C_{\color{green}{11}}$$ `=` $$\frac{\color{purple}{13}!}{(\color{purple}{13}-\color{green}{11})!\color{green}{11}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{13!}{2! 11!} $$ `=` $$ \frac{13\cdot12\cdot{\color{#CC0000}{11}}\cdot{\color{#CC0000}{10}}\cdot{\color{#CC0000}{9}}\cdot{\color{#CC0000}{8}}\cdot{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{2\cdot1\cdot{\color{#CC0000}{11}}\cdot{\color{#CC0000}{10}}\cdot{\color{#CC0000}{9}}\cdot{\color{#CC0000}{8}}\cdot{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}} $$ `=` $$ \frac{156}{2} $$ Cancel like terms `=` $$78$$ `78` -
Question 6 of 6
6. Question
How many ways can we form a `5` man basketball team from a group of `9` players?- (126)
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
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$We need to find the different ways that a team of `5` players `(r)` can be selected from a total of `9` players `(n)``r=5``n=9`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{9}C_{\color{green}{5}}$$ `=` $$\frac{\color{purple}{9}!}{(\color{purple}{9}-\color{green}{5})!\color{green}{5}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{9!}{4! 5!} $$ `=` $$ \frac{9\cdot8\cdot7\cdot6\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{4\cdot3\cdot2\cdot1\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}} $$ `=` $$ \frac{9\cdot{\color{#CC0000}{8}}\cdot7\cdot6}{{\color{#CC0000}{4}}\cdot3\cdot{\color{#CC0000}{2}}\cdot1} $$ Cancel like terms `=` $$ \frac{9\cdot7\cdot6}{3\cdot1} $$ `4xx2=8` `=` $$ \frac{378}{3} $$ `=` $$126$$ `126`
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