Combinations 2
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Question 1 of 6
1. Question
How many ways can students answer `6` questions on a `10` question quiz?- (210)
Hint
Help VideoCorrect
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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 questions `(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 students can answer `6` questions `(r)` on a `10` question quiz `(n)``r=6``n=10`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{10}C_{\color{green}{6}}$$ `=` $$\frac{\color{purple}{10}!}{(\color{purple}{10}-\color{green}{6})!\color{green}{6}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{10!}{4! 6!} $$ `=` $$ \frac{10\cdot9\cdot8\cdot7\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{4\cdot3\cdot2\cdot1\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}} $$ `=` $$ \frac{10\cdot9\cdot{\color{#CC0000}{8}}\cdot7}{{\color{#CC0000}{4}}\cdot3\cdot{\color{#CC0000}{2}}\cdot1} $$ Cancel like terms `=` $$ \frac{10\cdot9\cdot7}{3\cdot1} $$ `4xx2=8` `=` $$ \frac{630}{3} $$ `=` $$210$$ `210` -
Question 2 of 6
2. Question
How many ways can we form a group of `3` men and `4` women from a group of `5` men and `6` women?- (150)
Hint
Help VideoCorrect
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
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$First, we need to find the different ways that `4` women `(r)` can be selected from a total of `6` women `(n)``r=4``n=6`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{6}C_{\color{green}{4}}$$ `=` $$\frac{\color{purple}{6}!}{(\color{purple}{6}-\color{green}{4})!\color{green}{4}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{6!}{2! 4!} $$ `=` $$ \frac{6\cdot5\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{2\cdot1\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}} $$ `=` $$ \frac{6\cdot5}{2\cdot1} $$ Cancel like terms `=` $$ \frac{30}{2} $$ `=` $$15$$ Next, we need to find the different ways that `3` men `(r)` can be selected from a total of `5` men `(n)``r=3``n=5`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{5}C_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{5}!}{(\color{purple}{5}-\color{green}{3})!\color{green}{3}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{5!}{2! 3!} $$ `=` $$ \frac{5\cdot4\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{2\cdot1\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}} $$ `=` $$ \frac{5\cdot4}{2\cdot1} $$ Cancel like terms `=` $$ \frac{20}{2} $$ `=` $$10$$ Finally, multiply the values to get the total number of ways.`15``times``10` `=` `150` `150` -
Question 3 of 6
3. Question
How many ways can we form a group of `4` men and `3` women from a group of `7` men and `5` women?- (350)
Hint
Help VideoCorrect
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
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$First, we need to find the different ways that `3` women `(r)` can be selected from a total of `5` women `(n)``r=3``n=5`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{5}C_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{5}!}{(\color{purple}{5}-\color{green}{3})!\color{green}{3}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{5!}{2! 3!} $$ `=` $$ \frac{5\cdot4\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{2\cdot1\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}} $$ `=` $$ \frac{5\cdot4}{2\cdot1} $$ Cancel like terms `=` $$ \frac{20}{2} $$ `=` $$10$$ Next, we need to find the different ways that `4` men `(r)` can be selected from a total of `7` men `(n)``r=4``n=7`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{7}C_{\color{green}{4}}$$ `=` $$\frac{\color{purple}{7}!}{(\color{purple}{7}-\color{green}{4})!\color{green}{4}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{7!}{3! 4!} $$ `=` $$ \frac{7\cdot6\cdot5\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{3\cdot2\cdot1\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}} $$ `=` $$ \frac{7\cdot{\color{#CC0000}{6}}\cdot5}{{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot1} $$ Cancel like terms `=` $$ \frac{7\cdot5}{1} $$ `3xx2=6` `=` $$35$$ Finally, multiply the values to get the total number of ways.`10``times``35` `=` `350` `350` -
Question 4 of 6
4. Question
How many ways can a `12`-member panel be formed from a total pool of `38` people?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 `12`-member panel `(r)` can be formed from a pool of `38` people `(n)``r=12``n=38`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{38}C_{\color{green}{12}}$$ `=` $$\frac{\color{purple}{38}!}{(\color{purple}{38}-\color{green}{12})!\color{green}{12}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{38!}{26! 12!} $$ `=` `2 707 475 148` Use the calculator’s factorial function for large numbers `2 707 475 148` -
Question 5 of 6
5. Question
In how many ways can `4` cards be chosen from a standard deck of `52` cards?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
$$ _\color{purple}{n}C_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$We need to find the different ways that `4` cards `(r)` can be chosen from a standard deck of `52` cards `(n)``r=4``n=52`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{52}C_{\color{green}{4}}$$ `=` $$\frac{\color{purple}{52}!}{(\color{purple}{52}-\color{green}{4})!\color{green}{4}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{52!}{48! 4!} $$ `=` `270 725` Use the calculator’s factorial function for large numbers `270 725` -
Question 6 of 6
6. Question
In how many ways can `3` marbles be drawn from the jar below?
- (120)
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 total number of marbles (`n`)
`n` `=` `2` black `+3` red `+5` blue `n` `=` `10` marbles We need to find the different ways that `3` marbles `(r)` can be chosen from a jar of `10` marbles `(n)``r=3``n=10`$$_\color{purple}{n}C_{\color{green}{r}}$$ `=` $$ \frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!\color{green}{r}!} $$ Combination Formula $$_\color{purple}{10}C_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{10}!}{(\color{purple}{10}-\color{green}{3})!\color{green}{3}!} $$ Substitute the values of `r` and `n` `=` $$ \frac{10!}{7! 3!} $$ `=` $$ \frac{10\cdot9\cdot8\cdot{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}\cdot3\cdot2\cdot1} $$ `=` $$ \frac{10\cdot9\cdot8}{3\cdot2\cdot1} $$ Cancel like terms `=` $$ \frac{720}{6} $$ `=` $$120$$ `120`
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