Permutation Problems 2
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Question 1 of 5
1. Question
How many ways can `8` paintings be arranged in a row?- (40320)
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Use the permutations formula to find the number of ways an item can be arranged `(r)` from the total number of items `(n)`.Remember that order is important in Permutations.Permutation Formula
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$Fundamental Counting Principle
number of ways `=``m``times``n`Method OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each position in the rowFirst position:There are `8` paintings to choose from`=``8`Second position:From `8` paintings, one has already been chosen. Hence, we are left with `7` choices`=``7`Third position:From `8` paintings, two has already been chosen. Hence, we are left with `6` choices`=``6`Fourth position:From `8` paintings, three has already been chosen. Hence, we are left with `5` choices`=``5`Fifth position:From `8` paintings, four has already been chosen. Hence, we are left with `4` choices`=``4`Sixth position:From `8` paintings, five has already been chosen. Hence, we are left with `3` choices`=``3`Seventh position:From `8` paintings, six has already been chosen. Hence, we are left with `2` choices`=``2`Eighth position:From `8` paintings, seven has already been chosen. Hence, we are left with `1` choice`=``1`Use the Fundamental Counting Principle and multiply the number of options for each row.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `8``times``7``times``6``times``5``times``4``times``3``times``2``times``1` `=` `40 320` There are `40 320` ways to arrange `8` paintings`40 320`Method TwoWe are arranging `8` paintings `(r)` for `8` positions in a row `(n)``r=8``n=8`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{8}P_{\color{green}{8}}$$ `=` $$\frac{\color{purple}{8}!}{(\color{purple}{8}-\color{green}{8})!}$$ Substitute the value of `n` and `r` `=` $$\frac{8!}{0!}$$ `=` $$\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{1}$$ `0! =1` `=` `40 320` There are `40 320` ways to arrange `8` paintings`40 320` -
Question 2 of 5
2. Question
Out of `8` paintings, how many ways can `5` paintings be chosen to be arranged in a row?- (6720)
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Use the permutations formula to find the number of ways an item can be arranged `(r)` from the total number of items `(n)`.Remember that order is important in Permutations.Permutation Formula
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$Fundamental Counting Principle
number of ways `=``m``times``n`Method OneSolve the problem using the Fundamental Counting PrincipleFirst, count the options for each position in the rowFirst position:There are `8` paintings to choose from`=``8`Second position:From `8` paintings, one has already been chosen. Hence, we are left with `7` choices`=``7`Third position:From `8` paintings, two has already been chosen. Hence, we are left with `6` choices`=``6`Fourth position:From `8` paintings, three has already been chosen. Hence, we are left with `5` choices`=``5`Fifth position:From `8` paintings, four has already been chosen. Hence, we are left with `4` choices`=``4`Use the Fundamental Counting Principle and multiply each draw’s number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `8``times``7``times``6``times``5``times``4` `=` `6720` There are `6720` ways to arrange `5` paintings out of `8``6720`Method TwoWe are arranging `5` paintings `(r)` out of `8` paintings`(n)``r=5``n=8`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{8}P_{\color{green}{8}}$$ `=` $$\frac{\color{purple}{8}!}{(\color{purple}{8}-\color{green}{5})!}$$ Substitute the value of `n` and `r` `=` $$\frac{8!}{3!}$$ `=` $$\frac{8\cdot7\cdot6\cdot5\cdot4\cdot\color{#CC0000}{3\cdot2\cdot1}}{\color{#CC0000}{3\cdot2\cdot1}}$$ `=` `6720` Cancel like terms and evaluate There are `6720` ways to arrange `5` paintings out of `8``6720` -
Question 3 of 5
3. Question
In a box containing `8` cards marked `1`-`8`, what is the probability of drawing `345` in that order?Hint
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Probability
$$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$$Permutation Formula
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$First, find the total number of ways we can draw `3` cards `(r)` from a box containing `8` cards `(n)``r=3``n=8`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{8}P_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{8}!}{(\color{purple}{8}-\color{green}{3})!}$$ Substitute the values of `n` and `r` `=` $$\frac{8!}{5!}$$ `=` $$\frac{8\cdot7\cdot6\cdot\color{#CC0000}{5\cdot4\cdot3\cdot2\cdot1}}{\color{#CC0000}{5\cdot4\cdot3\cdot2\cdot1}}$$ `=` $$336$$ Cancel like terms and evaluate There are `336` ways to draw `3` cards from a box containing `8` cards. This is the total outcomeRemember that we want to draw the cards `345` in that order. This means that the favourable outcome is only `1`.Compute for the probability.Probability `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$$ `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{336}}$$ The probability of drawing the cards `345` in that order is `1/336``1/336` -
Question 4 of 5
4. Question
In a horse race with `13` competing horses, what is the probability of winning a Trifecta (successfully selecting the `1`st, `2`nd and `3`rd places)?
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Probability
$$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$$Permutation Formula
$$ _\color{purple}{n}P_{\color{green}{r}}=\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!} $$First, find the total number of ways we can select `3` horses `(r)` out of `13` `(n)``r=3``n=13`$$_\color{purple}{n}P_{\color{green}{r}}$$ `=` $$\frac{\color{purple}{n}!}{(\color{purple}{n}-\color{green}{r})!}$$ Permutation Formula $$_\color{purple}{13}P_{\color{green}{3}}$$ `=` $$\frac{\color{purple}{13}!}{(\color{purple}{13}-\color{green}{3})!}$$ Substitute the values of `n` and `r` `=` $$\frac{13!}{10!}$$ `=` $$\frac{13\cdot12\cdot11\cdot\color{#CC0000}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}{\color{#CC0000}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}}$$ `=` $$1716$$ Cancel like terms and evaluate There are `1716` ways to select `3` horses from `13` competing horses in the race. This is the total outcomeRemember that we want to successfully pick the `1`st, `2`nd and `3`rd ranking horse. This means that the favourable outcome is only `1`.Compute for the probability.Probability `=` $$\frac{\color{#e65021}{\mathsf{favourable\:outcome}}}{\color{#007DDC}{\mathsf{total\:outcome}}}$$ `=` $$\frac{\color{#e65021}{1}}{\color{#007DDC}{1716}}$$ The probability of a trifecta is `1/1716``1/1716` -
Question 5 of 5
5. Question
How many ways can we draw `3` cards from a box containing `8` cards marked with `1`-`8`, if each card is drawn with replacement?- (512)
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Fundamental Counting Principle
number of ways `=``m``times``n`First, count the options for each drawRemember that the cards are being drawn with replacementFirst to Third draw:There `8` cards in the box. Since each drawn card is replaced, all three draws have `8` options`=``8`Use the Fundamental Counting Principle and multiply each category’s number of options.number of ways `=` `m``times``n` Fundamental Counting Principle `=` `8``times``8``times``8` `=` `512` There are `512` ways of drawing `3` cards from the box with replacement`512`
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