Due to Corona Virus I lost my opportunity to attend in class, so I found this video very helpful. Thank you a lot. Btw. your voice reminds me of SsethTzeentach
@19:38 I don't think you are supposed to -4 there, you already subtracted 1 from 15 so you are essentially subtracting 5 instead there. Just think about 4 + 1 + 1 + 1 + 1 should be 8, 15-8 = 7, what happened was we did 5 + 1 + 1 + 1 + 1 to make 9, 15-9 = 6 This would also mean we get a 3rd situation where 4+4+4+1+2 = 15, which is definitely a possibility, so we have to subtract in (5 choose 3) (5 choose 1)
You have considered the universe (N) in the first question as 19C17 which shows the solution set is non-negative integers. I think that should be specified in the question or else it means find all possible solutions including negative integers.
Firstly, thank you so much for these videos, they make studying for my degree a lot more manageable and easy to understand :) I have encountered this problem on a past paper: In how many ways can 675 identical envelopes be divided, in packages of 25, among four student groups so that each group gets at least 150, but no more than 200, of the envelopes? I so far have that x1+x2+x3+x4=(?) such that 150
Hi! First off, thank you so much for your videos! I wouldn't have understood many concepts in my Discrete Structures course without them. I had a question regarding problem 1. Why do we add 3 and then subtract 1? Why do we need to subtract 1?
Hey! Bit late but in case anyone else had the same question, look at page 9 Theorem 1.11 of this link math.dartmouth.edu//archive/m68f07/public_html/lectec.pdf
Hi Trev, I really liked your video, although I just want to make a quick remark as it took me some time to understand it. In your first example, the notation you use "N(~c1,~c2,~c3)" is a little ambiguous. The way you used this notation in the rest of the exercise leads me to think that you meant "~c1 U ~c2 U ~c3" and if so, that it is wrong, as what we want is a conjunction of the conditions ~c1 & ~c2 & ~c3 (we want all the boxes to have less than or 7 balls, simultaneously), and that, by De Morgan's law, is the same as ~(c1 U c2 U c3), which in turn is the same as all the cases minus (c1 U c2 U c3).
Dear Trev Tutor, thank you so much for this videos! For Problem #1, we are assuming that we are talking about non-negative integers right? Also, how come sometimes Sn(the last condition that encompasses all conditions) be subtracted (like in the arrangements problem, or the flowers), but in the divisibility it's added!? (NEVER MIND I JUST SAW THAT AT THE END OF THE EXPLANATION YOU ACTUALLY SUBTRACTED Sn)
Thanks for all of your videos. But in this solution @ 19:38, I think x1'+x2'+x3'+x4'+x5'=7, not 6, because (x1+4)+(x2+1)+(x3+1)+(x4+1)+(x5+1)=15 will be considered. So instead of 10C6, I think it should be 11C7 {=(7+5-1)C7 }. The same concept may continue for S2 and S3 as well 20:26. So, S2 should be 5C2 * 8C4 instead of 5C2 * 6C2 and S3 should be 5C3 * 5C5 instead of 0 (zero). S3 can be non zero as we can keep 4 flowers in 3 shelves and 1 each in the other two (minimum requirement) and still left with 1 more to keep in any of the last two shelves. It would be great if you could kindly clarify.
i want to know the solution to this problem please: In how many ways can 26 alphabet letters be permuted so that none of the patterns car,dog,pun,or byte occurs? and thanks a lot.
+gehad mohsen I'll give an example for a couple, then explain the procedure for the rest. This is inclusion-exclusion, so we want Total-(One Word at a time)+(Two words at a time)-(Three words at a time)+(4 words at a time). There's 26! ways to do all the words. So we have 26! so far. There's 24! ways to compute a single word with 3 letters (car,b,d,e,f,g,h,i,j,k,l,m,n,o,p,q,s,t,u,v,w,x,y,z). There's 23! ways to compute a single word with 4 letters. So we have 26! - (3*24!*23!) so far. Now if we take car and pun together, we have 22! ways to permute the letters/words (car, pun, b,d,e,f,g,h,i,j,k,l,m,o,q,s,t,v,w,x,y,z) Then we continue for each pair/triplet/quadruplet of words.
Great answer, what if the word was PAN instead of PUN. do we have to do something special because the A is shown twice? Is it then 23! (one more than with CAR and PUN) or is it the same so also 22! ? Thanks!
Hi, I have s slightly similar problem to counting flowers and shelfs, but instead I have 8 red flowers and 10 yellow flowers and 4 shelfs and each shelf should have at least one flower (red or yellow). All 18 flowers need to be distributed across the 4 shelfs. What is comes to my mind: x1 + x2 + x3 + x4 = 18 where every x >= 1 => y = x + 1 y1 + y2 + y3 + y4 = 14 so total number of possible distributions will be C(17, 14) = 680 But do I have to take into account that there is 2 kinds of flowers (reds and yellows) or it doesn't matter in this situation? Regards.
mikeg What you've done there is the second step. Because there are two different colors of flowers, you have to figure out all the ways you can arrange those. So because we're dealing with two different objects we have 18!/(10!8!) because there are 10 yellows and 8 reds. Then we multiply by the number of ways to separate the flowers into shelves, which you have C(17,14). [THIS SOLUTION ASSUMES ORDER ON THE SHELF MATTERS] The best way to figure it out is to try smaller examples first. For instance, you can probably manually calculate 2 Blue, 2 Yellow on 2 shelves, and then figure out which method gets you the correct answer. Then you can extend it to use bigger numbers.
TheTrevTutor and if the order on the shelf DOESN'T matters, then I would have to divide the result by 4! (as the number of shelfs)? Can this problem be solved by using inclusion-exclusion principle as well? Thanks!
mikeg Things then get much, much complicated. I'd suggest taking that problem over math.stackexchange.com and tagging it in discrete-mathematics. They have LaTeX support there so the answers you get won't look like a garbled mess (which it will here, for sure. It's a lot more involved)
Hi. Great video. Only thing is I have problems with counting(for example when you counted N in the first problem). Any good books or videos on your cannel that you might recommend?
Seems like you're doing the exact same thing when x = 2, and when x >= 8.... How are you taking into account that when x >= 8, that x could be 9, or 10, or 11 etc?
An easier way to do the first problem is to substitue x1=7-a, x2=7-b, x3=7-c and then solve to get a+b+c=4 , which means x1,x2,x3 wont be negative so we know all combinations of a,b,c give us a working combination of x1,x2,x3 and the number of solutions to a+b+c=4 is 6C2=15 which is what your answer simplified to.
Hi, Nice video. Thanks a lot. I don't understand how the S0-S1+S2-S3+S4-S5 come from when you talked about the discretemath rocks example. Would hou please explain more for me. Thank you
4:53 Slipped some Einstein notation in there. Nice, but might want to explain for some viewers. Here's my little time-waster... DISCRETEMATHROCKS with no adjacent repeats has 9951856992000 arrangements.
![[Discrete Mathematics] Inclusion-Exclusion: At Least & Exactly](http://i.ytimg.com/vi/D1T3xy_vtxU/mqdefault.jpg)
![[Discrete Mathematics] Inclusion-Exclusion: At Least & Exactly](/img/tr.png)
Your explanation is more valuable than most of my professors'. They do nothing but complicate things. Thank you so much!
Thank you soooo much for these playlists! You literally just saved my grade.
Due to Corona Virus I lost my opportunity to attend in class, so I found this video very helpful. Thank you a lot.
Btw. your voice reminds me of SsethTzeentach
@19:38 I don't think you are supposed to -4 there, you already subtracted 1 from 15 so you are essentially subtracting 5 instead there. Just think about 4 + 1 + 1 + 1 + 1 should be 8, 15-8 = 7, what happened was we did 5 + 1 + 1 + 1 + 1 to make 9, 15-9 = 6
This would also mean we get a 3rd situation where 4+4+4+1+2 = 15, which is definitely a possibility, so we have to subtract in (5 choose 3) (5 choose 1)
@TheTrevTutor I also get this solution
S1 = (5 choose 1)(11 choose 7),
S2 = (5 choose 2)(7 choose 3) and so on...
Why my solution docent correct??
so this would be 10 choose 3 not 4 am i right? because we have 3 seperaters and 7 flowers
Best classes I've seen about this subject
Your videos are so helpful, youre saving my grades, tnx all the way from Israel!
I keep coming back to this channel. Thanks for DM videos.
Quick technical note on the last question: I believe you meant how many natural numbers n
You have considered the universe (N) in the first question as 19C17 which shows the solution set is non-negative integers. I think that should be specified in the question or else it means find all possible solutions including negative integers.
god of discrete mathematics. u have made things so simple. thanx a lot.
Amazing man ..! Literally watching it an hour before my final
Truly helpful 🔥
Firstly, thank you so much for these videos, they make studying for my degree a lot more manageable and easy to understand :)
I have encountered this problem on a past paper:
In how many ways can 675 identical envelopes be divided, in packages
of 25, among four student groups so that each group gets at least 150,
but no more than 200, of the envelopes?
I so far have that x1+x2+x3+x4=(?) such that 150
Emily Mcclymont You're close. So at first it seems like x1+x2+x3+x4=675 with 150
Thank you for the tutorials. Can you provide links for some more practice exercises with similar level of questions? Would be really helpful!
i hurt my leg so i couldn't attend to lessons. if i you wouldn't have done this awesome videos I was going to fail the course. thanks in advance
Thank you very much, Sir. I learn so much because of this channel
Q1: let x1=x2=x3=7, we'll arive at the solution by subtracting (3*7-17)=4
allocate 4*(-1) to 3 terms=6C4=15
done
Thanks and good joke with the s3
At 06:35, can you please explain why we choose 2 out of 3? I mean why did we do 3 chose 2 and not 3 chose 1?
3C1 and 3C2 evaluate to the same thing. Remeber, nCr is the same as nC(n-r)
very helpful for my 10th-grade math exam thanks!
U r brilliant man.Thanks a lot for this fabulous video.
Great video
I have a question though, at 15:32, are the flowers and shelves distinguishable? Thank you.
Your videos are really helpful. Thanks!
Hi! First off, thank you so much for your videos! I wouldn't have understood many concepts in my Discrete Structures course without them.
I had a question regarding problem 1. Why do we add 3 and then subtract 1? Why do we need to subtract 1?
Hey! Bit late but in case anyone else had the same question, look at page 9 Theorem 1.11 of this link math.dartmouth.edu//archive/m68f07/public_html/lectec.pdf
@@sobiakazmi4271 thanks
What a life saver you are , Not all Heroes Wear Capes!
shouldn't it be 19 choose 16? (n+k-1 choose k) = (n+k-1 choose n-1)
You are an awesome teacher.
Thanks
Great Tutorial. Thank you so much
Hi Trev, I really liked your video, although I just want to make a quick remark as it took me some time to understand it.
In your first example, the notation you use "N(~c1,~c2,~c3)" is a little ambiguous. The way you used this notation in the rest of the exercise leads me to think that you meant "~c1 U ~c2 U ~c3" and if so, that it is wrong, as what we want is a conjunction of the conditions ~c1 & ~c2 & ~c3 (we want all the boxes to have less than or 7 balls, simultaneously), and that, by De Morgan's law, is the same as ~(c1 U c2 U c3), which in turn is the same as all the cases minus (c1 U c2 U c3).
8:08 hahahah nice joke.. Made me laugh so bad.
Great sir a fab explanation to inclusion exclusion principle
Amazing Tutorial! Thank you!
Dear Trev Tutor, thank you so much for this videos!
For Problem #1, we are assuming that we are talking about non-negative integers right?
Also, how come sometimes Sn(the last condition that encompasses all conditions) be subtracted (like in the arrangements problem, or the flowers), but in the divisibility it's added!? (NEVER MIND I JUST SAW THAT AT THE END OF THE EXPLANATION YOU ACTUALLY SUBTRACTED Sn)
+Daniel Barrundia Nearly all counting problems will be with non-negative integers unless specified
you are the best
thanks for making my day
Thank you soo much! You're such a champ!
Thanks for all of your videos. But in this solution @ 19:38, I think x1'+x2'+x3'+x4'+x5'=7, not 6, because (x1+4)+(x2+1)+(x3+1)+(x4+1)+(x5+1)=15 will be considered. So instead of 10C6, I think it should be 11C7 {=(7+5-1)C7 }. The same concept may continue for S2 and S3 as well 20:26. So, S2 should be 5C2 * 8C4 instead of 5C2 * 6C2 and S3 should be 5C3 * 5C5 instead of 0 (zero). S3 can be non zero as we can keep 4 flowers in 3 shelves and 1 each in the other two (minimum requirement) and still left with 1 more to keep in any of the last two shelves. It would be great if you could kindly clarify.
+1
@15:32 Fifteen IDENTICAL flowers, that is.
Thanks. I was hoping someone said if they were identical or unique, before watching the solution!
thank you so much for this lesson..
for the flower pot problem, it should be C(14, 4) because you are arranging it into 5 distinct shelves and not 11
it's the same thing
Do you have a video for solving the equation if a lower limit is given for the variables?
i want to know the solution to this problem please:
In how many ways can 26 alphabet letters be permuted so that none of the patterns car,dog,pun,or byte occurs?
and thanks a lot.
+gehad mohsen I'll give an example for a couple, then explain the procedure for the rest.
This is inclusion-exclusion, so we want Total-(One Word at a time)+(Two words at a time)-(Three words at a time)+(4 words at a time).
There's 26! ways to do all the words.
So we have 26! so far.
There's 24! ways to compute a single word with 3 letters (car,b,d,e,f,g,h,i,j,k,l,m,n,o,p,q,s,t,u,v,w,x,y,z).
There's 23! ways to compute a single word with 4 letters.
So we have 26! - (3*24!*23!) so far.
Now if we take car and pun together, we have 22! ways to permute the letters/words (car, pun, b,d,e,f,g,h,i,j,k,l,m,o,q,s,t,v,w,x,y,z)
Then we continue for each pair/triplet/quadruplet of words.
Great answer, what if the word was PAN instead of PUN. do we have to do something special because the A is shown twice? Is it then 23! (one more than with CAR and PUN) or is it the same so also 22! ? Thanks!
@4:30 Why is it 11C9 should it not be 11 choose 3?
It is the same thing it's symmetrical you could say.
thank you so much for this
Hi,
I have s slightly similar problem to counting flowers and shelfs,
but instead
I have 8 red flowers and 10 yellow flowers and 4 shelfs and each shelf should have at least one flower (red or yellow). All 18 flowers need to be distributed across the 4 shelfs.
What is comes to my mind:
x1 + x2 + x3 + x4 = 18 where every x >= 1 => y = x + 1
y1 + y2 + y3 + y4 = 14
so total number of possible distributions will be C(17, 14) = 680
But do I have to take into account that there is 2 kinds of flowers (reds and yellows) or it doesn't matter in this situation?
Regards.
mikeg What you've done there is the second step. Because there are two different colors of flowers, you have to figure out all the ways you can arrange those. So because we're dealing with two different objects we have 18!/(10!8!) because there are 10 yellows and 8 reds. Then we multiply by the number of ways to separate the flowers into shelves, which you have C(17,14). [THIS SOLUTION ASSUMES ORDER ON THE SHELF MATTERS]
The best way to figure it out is to try smaller examples first. For instance, you can probably manually calculate 2 Blue, 2 Yellow on 2 shelves, and then figure out which method gets you the correct answer. Then you can extend it to use bigger numbers.
TheTrevTutor and if the order on the shelf DOESN'T matters, then I would have to divide the result by 4! (as the number of shelfs)?
Can this problem be solved by using inclusion-exclusion principle as well?
Thanks!
mikeg Things then get much, much complicated. I'd suggest taking that problem over math.stackexchange.com and tagging it in discrete-mathematics. They have LaTeX support there so the answers you get won't look like a garbled mess (which it will here, for sure. It's a lot more involved)
TheTrevTutor Thanks, I'll try to ask there!
استمر يرايق
very helpful video.thank u sir
Hi. Great video. Only thing is I have problems with counting(for example when you counted N in the first problem). Any good books or videos on your cannel that you might recommend?
What is the floor function and ceiling function?
In DISRCETEMATHROCKS problem , dont you think in S2, EERR or RREE will be 2 permutations so we should multiply it by 2 ?
Seems like you're doing the exact same thing when x = 2, and when x >= 8.... How are you taking into account that when x >= 8, that x could be 9, or 10, or 11 etc?
An easier way to do the first problem is to substitue x1=7-a, x2=7-b, x3=7-c and then solve to get a+b+c=4 , which means x1,x2,x3 wont be negative so we know all combinations of a,b,c give us a working combination of x1,x2,x3 and the number of solutions to a+b+c=4 is 6C2=15 which is what your answer simplified to.
Can someone explain why it’s s1 plus s2 and not minus on all in the discretemathrocks problem? :) thanks
Thank u so much sir
thanks for the video
Hi, Nice video. Thanks a lot. I don't understand how the S0-S1+S2-S3+S4-S5 come from when you talked about the discretemath rocks example. Would hou please explain more for me. Thank you
Because of no of repated letters (ss cc rr ee tt)
for the last question, why are the conditions 3/n 5/n and 7/n
amazing vid
4:53 Slipped some Einstein notation in there. Nice, but might want to explain for some viewers.
Here's my little time-waster... DISCRETEMATHROCKS with no adjacent repeats has 9951856992000 arrangements.
Are you an Olympian?
Give each person 7, Now you have given 21 so you need to take back 4. i.e. x1 + x2 + x3 = 4. so you get 6C2
How you got 11??
Thx
HELP TO EVERYONE
Who else CS major?
Hahaha that k joke made me cry
Cool
i love you