Combinations 1

Question 1 of 6

Evaluate

$_4 C_3$

(4)

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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 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

Evaluate

$_{10} C_4$

(210)

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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 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

Evaluate

$_9 C_2$

(36)

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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 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

How many ways can we pick

$5$ paintings from a gallery with

$14$ paintings?

(2002)

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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 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

How many ways can a soccer team of

$11$ players be selected from a squad of

$13$ existing players?

1. $156$
2. $11$
3. $89$
4. $78$

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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 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

How many ways can we form a

$5$ man basketball team from a group of

$9$ players?

(126)

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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 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

	$=$	$\frac{9\cdot7\cdot6}{3\cdot1}$	$4xx2=8$

	$=$	$\frac{378}{3}$

	$=$	$126$

$126$

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Well Done!

Combination Formula

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Excellent!

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Fantastic!

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Correct!

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