[Discrete Mathematics] Factorials and Permutations
ฝัง
- เผยแพร่เมื่อ 6 ต.ค. 2024
- Visit my website: bit.ly/1zBPlvm
Subscribe on TH-cam: bit.ly/1vWiRxW
Hello, welcome to TheTrevTutor. I'm here to help you learn your college courses in an easy, efficient manner. If you like what you see, feel free to subscribe and follow me for updates. If you have any questions, leave them below. I try to answer as many questions as possible. If something isn't quite clear or needs more explanation, I can easily make additional videos to satisfy your need for knowledge and understanding.
Today we introduce factorials, permutations, and permutations without repetition. We also do a few practice problems, but the next video will consist of only practice problems.
![[Discrete Mathematics] Subsets and Power Sets](http://i.ytimg.com/vi/9OWdaaEE0L4/mqdefault.jpg)
![[Live] : ONE FIGHT NIGHT 25 วันนี้!! "อเล็กซิส vs รีเกียน"](http://i.ytimg.com/vi/suVr9a6BM2g/mqdefault.jpg)
Well I started too late to study because of real-life. Exams seam to be lightining rod for family illnesses, so yep, they were ill.
and I scrapped a pass mark all down to your videos! THANK YOU!!!
I've been refreshing all my discete 1 math as I'm taking a 200 level course next month - I gotta say, your videos are fantastic. Keep doing what you're doing man. Decent pace, extraordinarily clear delivery, also - written on black is easy to read and easy on the eyes... and the voice over format is great. I'll be thumbing up every video.
some number over 840 got me like XD
+Devun Schmutzler hehe classic. 36 x 840 = some number > 840
Wouldn't the work for "How many ways can we list 7 numbers without repetition?" be 10 * 9 * 8 * 7 * 6 * 6 * 4? Considering you have 10 options for digits in the beginning, 9 after choosing 1, 8 after choosing 2, etc. I mean I guess the video would be right if you defined "numbers" to just be the first 7 digits, but I think there should be some clarification to the question lol
Yeah, I should have specified the first 7 digits.
You are right
some number>840 You are great. Thanx for this video sir
who the fuck disliked
Roughly some number > 840 - HA! Who knew that discrete math could be so freakin' hilarious!
The entrance exam for an university is formed from 4 subjects. For each subject the student can get from 1 to 5 points. The student has passed the exam if he has at least 16 points and on each subject he has at least 20% from the maximum points(5).
In how many different ways the student can get enough points to pass the entrance exam?
I plugged it into Python and the answer is 70 possible ways. Not to sure on how to get there without brute force, though :/
1 : 1 5 5 5 total = 16
2 : 2 4 5 5 total = 16
3 : 2 5 4 5 total = 16
4 : 2 5 5 4 total = 16
5 : 2 5 5 5 total = 17
6 : 3 3 5 5 total = 16
7 : 3 4 4 5 total = 16
8 : 3 4 5 4 total = 16
9 : 3 4 5 5 total = 17
10 : 3 5 3 5 total = 16
11 : 3 5 4 4 total = 16
12 : 3 5 4 5 total = 17
13 : 3 5 5 3 total = 16
14 : 3 5 5 4 total = 17
15 : 3 5 5 5 total = 18
16 : 4 2 5 5 total = 16
17 : 4 3 4 5 total = 16
18 : 4 3 5 4 total = 16
19 : 4 3 5 5 total = 17
20 : 4 4 3 5 total = 16
21 : 4 4 4 4 total = 16
22 : 4 4 4 5 total = 17
23 : 4 4 5 3 total = 16
24 : 4 4 5 4 total = 17
25 : 4 4 5 5 total = 18
26 : 4 5 2 5 total = 16
27 : 4 5 3 4 total = 16
28 : 4 5 3 5 total = 17
29 : 4 5 4 3 total = 16
30 : 4 5 4 4 total = 17
31 : 4 5 4 5 total = 18
32 : 4 5 5 2 total = 16
33 : 4 5 5 3 total = 17
34 : 4 5 5 4 total = 18
35 : 4 5 5 5 total = 19
36 : 5 1 5 5 total = 16
37 : 5 2 4 5 total = 16
38 : 5 2 5 4 total = 16
39 : 5 2 5 5 total = 17
40 : 5 3 3 5 total = 16
41 : 5 3 4 4 total = 16
42 : 5 3 4 5 total = 17
43 : 5 3 5 3 total = 16
44 : 5 3 5 4 total = 17
45 : 5 3 5 5 total = 18
46 : 5 4 2 5 total = 16
47 : 5 4 3 4 total = 16
48 : 5 4 3 5 total = 17
49 : 5 4 4 3 total = 16
50 : 5 4 4 4 total = 17
51 : 5 4 4 5 total = 18
52 : 5 4 5 2 total = 16
53 : 5 4 5 3 total = 17
54 : 5 4 5 4 total = 18
55 : 5 4 5 5 total = 19
56 : 5 5 1 5 total = 16
57 : 5 5 2 4 total = 16
58 : 5 5 2 5 total = 17
59 : 5 5 3 3 total = 16
60 : 5 5 3 4 total = 17
61 : 5 5 3 5 total = 18
62 : 5 5 4 2 total = 16
63 : 5 5 4 3 total = 17
64 : 5 5 4 4 total = 18
65 : 5 5 4 5 total = 19
66 : 5 5 5 1 total = 16
67 : 5 5 5 2 total = 17
68 : 5 5 5 3 total = 18
69 : 5 5 5 4 total = 19
70 : 5 5 5 5 total = 20
So there is a difference when the question asks for permutations or a combinations.
Permutation implies no repetition and Combination implies repetition is allowed (if not stated otherwise)?
Having Dyslexia and assigning concepts and ideas to letters (and the combination of letters) is the harder part for me :D
+Paul Dixon Permutations can have repetition or not have repetition. Combinations can have repetition or not repetition. It will usually say in the question whether repetition is allowed or not.
It's up to you to figure out if it's combinations or permutations.
U HELP ME ALOT
This is actually cool
How did u write RRUUU!? Right Right, Up, Up UP!? @16:00
What is P() for BALL and DATABASES?
what if the question gives you 2 words like for ex: let's say BACK and FORTH, how can we find different strings that can be produced as shuffles of the following words?
Thank you so much
Outstanding
Are those videos based on a specific book?
abc,acb,cba,cab,bac,bca
aaa,aab,aba,baa,
bbb,bba,bab,abb,
ccc,cca,cac,acc,
ccb,cbc,bcc,cca,
caa,acc,aac,aca,
26 possible permutations,(if I didn't miss any)
How do I get to this?given 3 numbers in an array to make a list with all the possible permutations between them
27. 3 to the power of 3, if repetition is allowed.
I still get that till (4x3x2) which is the first three numbers are odd. How did the rest 4 number became (4x3x2x1). Rather than ( 7x6x5x4) , why it became another (4x3x2x1)? By how? There are 7 digit in the set. After the first three digit is odd, did the rest also need to be odd. Or else , there are 7digit which is it should be from 1 to 7 possibilities. With that, the rest number should be ( 7x6x5x4)
The question specifically says number shouldnt be repeated so the 'rest' cannot be odd cause they're already used
On the question regarding the DATABASES. You only made the 8 divided by 2!, what about 4 then? Because in the numerator the numbers were( 9*8*7*6*5*4)/2! and you only divided the 8 by 2!. Why didn't you divide the 4 by 2! too? Or is there's a rule that we can only divide 1 variable for the given factorial?
2!=2x1=2, so he just chose one term from the numerator to divide by 2 in order to eliminate the 2 from the denominator. Dividing the 4 by 2 as well would have caused the expression to have been divided by 4 total, so that's why he only did it with one term.
Okay so basically you have to divide one term if we have more than 1 to get rid of the denominator right?
Yeah, just divide one term from the top for each term in the bottom (as long as the terms for each are being multiplied and not added/subtracted)
Okay got it thank you :)
Here's one that bug me
I have 2 dogs and 10 cats. How many ways their is to align 5 animals (out of the 12) for a photo shot if
a) at least one dog has to be on the picture
b) The 2 dogs needs to be on the picture
+JP Dubé a) Break it into cases. How many ways with Dog 1? How many ways with Dog 2? How many ways with both Dog 1 and Dog 2?
b) 2 dogs are in the picture. So first you pick 2 out of the 2 dogs (2 choose 2), then pick 3 out of the 10 cats (10 choose 3), then order them.
savior bhai
For the: list 7 numbers but the first 3 have to be odd, why do we only have 4 odd numbers? If we can select any number fro 0-10, we have 5 odd numbers to choose from. What am I missing?
Hey, i had the same issue. However, i realized that the set of numbers he deals is between 1-7 (not 1-9 #10). So that leaves you with an option for the first three of odd numbers to be 1, 3, 5, 7. (4 options), then 3 options and 2 options.
From then on, he multiplies by the remaining 4, 3 and 2 options left over. Hope this helps :)
I am really confused here. I can not apply the n and k equation you provided to most of the cases in the practice section. Why do BALL has 4！/2！ permutations, instead of 4！/1! if there are 3 available alternatives out of 4. (n-k!) in this case should equal to (4-3)!.
+Cai Xinxin nvm i got it
Can you enlighten me?
uko sawa sisemi kitu
12:19 ......you got me 😂👑
I dont like how you didnt use the formula in the second part of the video. At least explain why you didn't. Or if you can then show how do it using the formula
Yup he didn't use the formula for the second part. It's almost like he was trying to show off.
It's a lot better to know what's actually going on than having to rely on a formula the whole time. That's the problem with educational systems that just throw formulas at students.
Mr. Trevor, also, if it's not much of a trouble, Could you quickly explain to me this one:
imgur.com/O26fhR9
I have to choose nPk, where k are the indices in which values are different. That's cool. But Why do I have to multiply the result of that permutation by: *2^k* and by: *2^n-k*?
Thank you!
Hello Trevor, Great video as usual, Could you please explain to me Why is this exercise: imgur.com/Goeu7Zj
solved in that way? I mean, why do you have to choose the groups of *20!*?
I understand that in this case every coder is undistinguishable, therefore *Why is it not just 97!/97!* the right solution?
Thank in advance!
JoseWanKenobi Consider 5 rooms: A, B, C, D, and E. I'm going to shrink this problem down a bit.
Let's say we want 4 coders in room A, B, C, D, and 3 coders in room E. Then we can assign them each a letter. Then we get a word like AAAABBBBCCCCDDDDEEE. So we just want to find all possible arrangements of the letters in AAAABBBBCCCCDDDDEEE, which is 19!/(4!4!4!4!3!). Same reasoning for your problem. (The first position is person 1, the second is person 2, etc.)
In other words, this is the same type of problem as finding the number of ways as rearranging the letters in ARRANGEMENT, which is covered in here, I think.
TheTrevTutor Fantastic Explanation, Thank you very much! Seeing the problem in the terms you put it is the key. Man, you're incredible, never stop doing videos.
I just don't get why we see the 2 L's as the same thing ? why can't we pretend they're different.
Because if I just type the word "HILL" to you, how do you know which L is which? It doesn't matter which L is which L, which is why we count them as the "same thing".
thanks that was a perfect answer.
CIS2910 anyone?
yeah yeah yeah, i know
Your videos are good but too long. I'd appreciate it more if you could go straight to the point and not waste time talking about irrelevant things. Also, you don't need to spend time doing the calculator's work.
The alternative is hour long lectures. I'm not sure how you can complain unless you are cramming.... You can also fast forward 5 seconds using your arrow keys
Protip: gear icon>speed>2
Lmao, this dude condenses 3 hour long lecture topics into 16 minutes and you're complaining that they're too long? If anything, I wish they were longer. There's literally no fat on these things, so I can't see where you're getting "irrelevant things" from.