Permutations with Repetition | Combinatorics, Permutations and Combinations
36.8 هزار بار بازدید -
5 سال پیش
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How do we count permutations
How do we count permutations when repetition is allowed? Problems like counting how many 4-digit codes can be created with the digits 0-9 - allowing repetition of digits - cannot be solved using the permutation formula. As it turns out, the permutation formula is more complicated than anything we need to solve the sort of problems we’re talking about today. We’ll be talking about permutations with repetition in today’s combinatorics video lesson!
You might have seen a more standard permutation problem where repetition isn’t allowed, and if you haven’t I recommend checking out my video on the permutation formula and why it works, where we go over some great examples and explanations! The Permutation Formula and Why it Wo...
For example, how many 4-digit codes can be created using the digits 0-9 without repetition? There are 10 choices for the first digit, but repetition is not allowed, so there are only 9 options for the second digit, 8 for the third, and 7 for the fourth. Thus, there is a total of 10*9*8*7 possible 4-digit codes in this situation.
What if repetition is allowed? Then there are 10 options for all 4 digits! Thus there are 10*10*10*10 = 10^4 total options. So if we have a collection of n objects, and are ordering r of them, and repetition is allowed, then there is a total of n^r permutations possible! Check out the full video lesson for more details and examples!
SOLUTION TO PRACTICE PROBLEM:
In this problem, the collection size is 26, there are 26 letters to choose from. We are ordering 7 of them because we are counting the number of 7 letter passwords. Repetition of letters is allowed. There is a total of 26^7 permutations of 7 letters from a collection of 26 when repetition is allowed. Simple as that, 26^7 is the answer!
If repetition was not allowed, the answer could be found using the permutation formula P(26, 7) = 26*25*24*23*22*21*20 = 26!/19!.
I uploaded this video a couple of days ago, but didn't like how I explained one thing, so I fixed it, and took down the old one!
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The outro music is by a favorite musician of mine named Vallow, who, upon my request, kindly gave me permission to use his music in my outros. I usually put my own music in the outros, but I love Vallow's music, and wanted to share it with those of you watching. Please check out all of his wonderful work.
Vallow Bandcamp: https://vallow.bandcamp.com/
Vallow Spotify: https://open.spotify.com/artist/0fRtu...
Vallow SoundCloud: SoundCloud: benwatts-3
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+WRATH OF MATH+
◆ Support Wrath of Math on Patreon: Patreon: wrathofmathlessons
Follow Wrath of Math on...
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You might have seen a more standard permutation problem where repetition isn’t allowed, and if you haven’t I recommend checking out my video on the permutation formula and why it works, where we go over some great examples and explanations! The Permutation Formula and Why it Wo...
For example, how many 4-digit codes can be created using the digits 0-9 without repetition? There are 10 choices for the first digit, but repetition is not allowed, so there are only 9 options for the second digit, 8 for the third, and 7 for the fourth. Thus, there is a total of 10*9*8*7 possible 4-digit codes in this situation.
What if repetition is allowed? Then there are 10 options for all 4 digits! Thus there are 10*10*10*10 = 10^4 total options. So if we have a collection of n objects, and are ordering r of them, and repetition is allowed, then there is a total of n^r permutations possible! Check out the full video lesson for more details and examples!
SOLUTION TO PRACTICE PROBLEM:
In this problem, the collection size is 26, there are 26 letters to choose from. We are ordering 7 of them because we are counting the number of 7 letter passwords. Repetition of letters is allowed. There is a total of 26^7 permutations of 7 letters from a collection of 26 when repetition is allowed. Simple as that, 26^7 is the answer!
If repetition was not allowed, the answer could be found using the permutation formula P(26, 7) = 26*25*24*23*22*21*20 = 26!/19!.
I uploaded this video a couple of days ago, but didn't like how I explained one thing, so I fixed it, and took down the old one!
********************************************************************
The outro music is by a favorite musician of mine named Vallow, who, upon my request, kindly gave me permission to use his music in my outros. I usually put my own music in the outros, but I love Vallow's music, and wanted to share it with those of you watching. Please check out all of his wonderful work.
Vallow Bandcamp: https://vallow.bandcamp.com/
Vallow Spotify: https://open.spotify.com/artist/0fRtu...
Vallow SoundCloud: SoundCloud: benwatts-3
********************************************************************
+WRATH OF MATH+
◆ Support Wrath of Math on Patreon: Patreon: wrathofmathlessons
Follow Wrath of Math on...
● Instagram: Instagram: wrathofmathedu
● Facebook: Facebook: WrathofMath
● Twitter: Twitter: wrathofmathedu
My Music Channel: seanemusic
5 سال پیش
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