Combinations 2

Question 1 of 6

How many ways can students answer

6

$6$ questions on a

10

$10$ question quiz?

(210)

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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 questions $(n)$ $(n)$ .

Remember that order is not important in Combinations.

Combination Formula

_{n} C_{r} = \frac{n!}{(n - r)! r!}

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

We need to find the different ways that students can answer $6$ $6$ questions $(r)$ $(r)$ on a $10$ $10$ question quiz $(n)$ $(n)$

r = 6

$r=6$

n = 10

$n=10$

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

	$=$ $=$	$\frac{10!}{4! 6!}$

	$=$ $=$	$\frac{10\cdot9\cdot8\cdot7\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{4\cdot3\cdot2\cdot1\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$

	$=$ $=$	$\frac{10\cdot9\cdot{\color{#CC0000}{8}}\cdot7}{{\color{#CC0000}{4}}\cdot3\cdot{\color{#CC0000}{2}}\cdot1}$	Cancel like terms

	$=$ $=$	$\frac{10\cdot9\cdot7}{3\cdot1}$	$4 \times 2 = 8$ $4xx2=8$

	$=$ $=$	$\frac{630}{3}$

	$=$ $=$	$210$

210

$210$

Question 2 of 6

How many ways can we form a group of

3

$3$ men and

4

$4$ women from a group of

5

$5$ men and

6

$6$ women?

(150)

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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, we need to find the different ways that $4$ $4$ women $(r)$ $(r)$ can be selected from a total of $6$ $6$ women $(n)$ $(n)$

r = 4

$r=4$

n = 6

$n=6$

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

	$=$ $=$	$\frac{6!}{2! 4!}$

	$=$ $=$	$\frac{6\cdot5\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{2\cdot1\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$

	$=$ $=$	$\frac{6\cdot5}{2\cdot1}$	Cancel like terms

	$=$ $=$	$\frac{30}{2}$

	$=$ $=$	$15$

Next, we need to find the different ways that $3$ $3$ men $(r)$ $(r)$ can be selected from a total of $5$ $5$ men $(n)$ $(n)$

r = 3

$r=3$

n = 5

$n=5$

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

	$=$ $=$	$\frac{5!}{2! 3!}$

	$=$ $=$	$\frac{5\cdot4\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{2\cdot1\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$

	$=$ $=$	$\frac{5\cdot4}{2\cdot1}$	Cancel like terms

	$=$ $=$	$\frac{20}{2}$

	$=$ $=$	$10$

Finally, multiply the values to get the total number of ways.

$15$ $15$

\times

$times$ $10$ $10$

=

$=$

150

$150$

150

$150$

Question 3 of 6

How many ways can we form a group of

4

$4$ men and

3

$3$ women from a group of

7

$7$ men and

5

$5$ women?

(350)

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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, we need to find the different ways that $3$ $3$ women $(r)$ $(r)$ can be selected from a total of $5$ $5$ women $(n)$ $(n)$

r = 3

$r=3$

n = 5

$n=5$

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

	$=$ $=$	$\frac{5!}{2! 3!}$

	$=$ $=$	$\frac{5\cdot4\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{2\cdot1\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}$

	$=$ $=$	$\frac{5\cdot4}{2\cdot1}$	Cancel like terms

	$=$ $=$	$\frac{20}{2}$

	$=$ $=$	$10$

Next, we need to find the different ways that $4$ $4$ men $(r)$ $(r)$ can be selected from a total of $7$ $7$ men $(n)$ $(n)$

r = 4

$r=4$

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

$_\color{purple}{7}C_{\color{green}{4}}$	$=$ $=$	$\frac{\color{purple}{7}!}{(\color{purple}{7}-\color{green}{4})!\color{green}{4}!}$	Substitute the values of $r$ $r$ and $n$ $n$

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

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

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

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

	$=$ $=$	$35$

Finally, multiply the values to get the total number of ways.

$10$ $10$

\times

$times$ $35$ $35$

=

$=$

350

$350$

350

$350$

Question 4 of 6

How many ways can a

12

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

38

$38$ people?

1. $2 707 475 148$ $2 707 475 148$
2. $3 145 475$ $3 145 475$
3. $5 103 724$ $5 103 724$
4. $707 475 148$ $707 475 148$

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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!}

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

We need to find the different ways that a $12$ $12$ -member panel $(r)$ $(r)$ can be formed from a pool of $38$ $38$ people $(n)$ $(n)$

r = 12

$r=12$

n = 38

$n=38$

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

	$=$ $=$	$\frac{38!}{26! 12!}$

	$=$ $=$	$2 707 475 148$ $2 707 475 148$	Use the calculator’s factorial function for large numbers

2 707 475 148

$2 707 475 148$

Question 5 of 6

In how many ways can

4

$4$ cards be chosen from a standard deck of

52

$52$ cards?

1. $1 340 829$ $1 340 829$
2. $455$ $455$
3. $270 725$ $270 725$
4. $384 125$ $384 125$

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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!}

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

We need to find the different ways that $4$ $4$ cards $(r)$ $(r)$ can be chosen from a standard deck of $52$ $52$ cards $(n)$ $(n)$

r = 4

$r=4$

n = 52

$n=52$

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

	$=$ $=$	$\frac{52!}{48! 4!}$

	$=$ $=$	$270 725$ $270 725$	Use the calculator’s factorial function for large numbers

270 725

$270 725$

Question 6 of 6

In how many ways can

3

$3$ marbles be drawn from the jar below?

(120)

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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 total number of marbles ( $n$ $n$ )

$n$ $n$	$=$ $=$	$2$ $2$ black $+ 3$ $+3$ red $+ 5$ $+5$ blue
$n$ $n$	$=$ $=$	$10$ $10$ marbles

We need to find the different ways that $3$ $3$ marbles $(r)$ $(r)$ can be chosen from a jar of $10$ $10$ marbles $(n)$ $(n)$

r = 3

$r=3$

n = 10

$n=10$

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

	$=$ $=$	$\frac{10!}{7! 3!}$

	$=$ $=$	$\frac{10\cdot9\cdot8\cdot{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}}{{\color{#CC0000}{7}}\cdot{\color{#CC0000}{6}}\cdot{\color{#CC0000}{5}}\cdot{\color{#CC0000}{4}}\cdot{\color{#CC0000}{3}}\cdot{\color{#CC0000}{2}}\cdot{\color{#CC0000}{1}}\cdot3\cdot2\cdot1}$

	$=$ $=$	$\frac{10\cdot9\cdot8}{3\cdot2\cdot1}$	Cancel like terms

	$=$ $=$	$\frac{720}{6}$

	$=$ $=$	$120$

120

$120$

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Great Work!

Combination Formula

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Combination Formula

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Combination Formula

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Combination Formula

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Combination Formula

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