Permutations with restrictions - items not together | ExamSolutions
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- เผยแพร่เมื่อ 22 พ.ค. 2011
- Permutations with restrictions.
In this tutorial I demonstrate how to calculate permutations (arrangements) where there are restrictions in place. In these examples certain items are not to be placed together. The examples used are:
1) In how many ways can 5 men and 3 women be arranged in a row if no two women are standing next to each other.
2) In how many ways can the letters in the word SUCCESS be arranged if no two S's are next to each other.
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idk what i did whole year now i finally know the concept of permutation ... all teachers in class were useless but you sir, are a legend :D
+marsal sarma I'm pleased to hear you have got it now.
+marsal sarma
yes u r write i also get a concept from that lecture great sir
same with me here haha
ExamSolutions sir, u know my teacher made my class write down all the possible arrangements everytime. After watching ur video, I finally understood why. She thought we were stupid =_=. BUT.... I UNDERSTOOD UR VIDEO. YEAAAA MANNN
This is what happens when you pay more attention to a youtube video than you payed to your teacher for a whole year.
God bless this man! explained way better than my teacher.
absolute legend! I have an awful stats teacher, but you have renewed my confidence in the subject. good stuff
Thank u so much sir. Actually you are a good teacher. I didn't really understand this lesson before but now I can understand because all of your Permutation lessons. Thank u so much again
This was the hardest part of the CIE AS level Maths Statistics as it was very hard to comprehend why u could draw spaces between and apply the permutation and combination formula whilst believing that the n of n! or nCr is a fixed total number of a person or item
Your method is correct.. sorry for doubting you I finally know why the other method doesn't work.
THANK GOD THIS LEGEND EXIST!!! THANKS MAN IT HELPED A LOT
Even in the second problem, we are obtaining a totally different answer by subtracting the number of ways where s' are together from the total number of ways:
SUCCESS:
Total arrangements possible= 7! / (3! . 2!)
Number of arrangements where s' are together = 5! / 2!
(Note: as no internal arrangement possible between s')
Number of ways where s' are not together = 420-60 =360.
Hi there im not sure if youre still active but i was curious, for the last question isnt it possible to have more permutations considering U And E are next to each other. S could still be separated in that case. As well as when C1 and C2 are together, or U and E are all the way at the edge. Theres just a lot more permutations other than the one shown that allows for S’s to be separated.
For instance
_ U E _C1_C2_
The S’s would still have spots separated here
In this case how do we consider all those other arrangements?
I know am the newest here .. the only thing that matter the 12 years old video helps ..old is gold great job
Some Help? Are you jokin', Sir? This was more than 'some help'. You cleared my misunderstandings. Thank you for that. Hats off :')
never mind.. i got it already, just need to watch your video for the 2nd time :D thanks your video helps me alot.
This is awesome keep up the good work ❤
Thank you!!! This was extremely helpful!!
BRO..... YOURE A LEGEND
is this all we have for As level stats in permutations and combinations?
Thank you very much... Our teacher at school told us that we should get the number of ways of arrangements when their are no restrictions and subtract it from when particular group of people are together in order for us to find the number of ways of arranging them when they are not next to each other...but why didn't this method work on the question in this video?
So what about questions which have more than 2 things which aren't together, does this method work for that too
Can't thank you enough for this! made the chapter look so simple
Always happy to help.
@@ExamSolutions_Maths I have a really important question to ask
thanks man! this was really helpful
@ExamSolutions i hope u know that ur the man and a chad
thank you so much u got a special place in heaven sir
Great tutorial, thanks a lot!
sir plz solve = in how many ways can eight people be seated in a row of eight seats so thet two particular persons are always together ? how 10080 ans ?
Ooooooof Jaaaaaniiii maza agaya Thank you soo much Boss!
Thank you so much .feel so lucky after i found this
Thank you so much for the videos sir. They are incredibly helpful for me especially since that I am studying the topics by myself. I have a doubt and I will really appreciate if you could help me out with it.
For the first question, shouldn't the answer be multiplied by 3! because the 3 women can arrange themselves differently?
You probably don't need this anymore, but in case anyone else gets confused, it should be multiplied by 6P3 (which is 6×5×4) because there are 6 spaces in which the 3 women can arrange themselves.
can we use the same method if the number of spaces are specified in the question? And thanks!
Sir,
based on the last problem regarding 'SUCCESS' string my problem is "number of arrangements that can be made out of the word , so that all S's do not come together is
some where i see answer as 360. Please explain
Thanks
Anes
great explanation thankssss
Thanks a lot sir you are the best🔥🔥🔥
Ur a great teacher. Appreciate u.
how about "eight students to be arranged in the row, ho many ways to arranged them if three particular student must be separated?
pls why did you divide the 3 factorial in the 2nd ques but in the first ques you didnt divide by 3 factorial
Damn, this works like a charm tq sir
How do you figure out how many dashes to give before and after the numbers?
you dont - you just need to sufficiently prepare the spaces in order to sub the elements positions inside - according to the question
Hey I was wondering if you could help me out with this question;
thank youuu so much this helped sooooo muchhhhhhhhhh!!
Thanks a lot, good sir. Appreciate that :)
Thank you Excellent explanation
very Well done
Nice job sir
this really helps me! thank u so much
Glad it helps
thank you, sir, for this
thank you again sir! 😇
Thanks , it really helps me
but , can we not solve it in 5! x 3! ways ??, please reply
for 2:13 why do we put space in between for M1 M2 M3 M4 M5
wow this is great
I really don't get how this works. You have 8 possible spaces, but you have made 10. Why could the men not stand next to each other in 1 or 2 instances. The conditions would still be satisfied.
I'm probably being really stupid, but that made no sense to me whatsoever.
They do. In the case a woman does not occupy the blank space, the two men would stand next to each other.
They would only not stand next to each other if a woman is standing between them :)
But can't the men be arranged close to one another right
why did u put 11 spaces?
I don't understand why for the boys it 5! Aren't there more than 5 ways for let's say the first guy to be placed somwhere?
Thanks man
You did not consider M1_M2_M3_M4_M5_ which would be 5! for the men and 5x4x3 for the women giving 5!x5x4x3=7200 and then why not W1_ _ _ W2 _ W3 _ or W1 _ _ W2 _ _ W3 _ and so on. You are not considering all possibilities. My method is a standard method for this type of question.
Thanks!!! Needed it for CPT.
Shree Kirloskar Cheers
Thanks, chap.
Thank you so muchh...
Well done
how about "example"? the two letters of e are not next to each other. ive tried your method but its wrong. tq
thank you so much sure awesome
Thanks 👍👍👍
What would I do without you sir?
Thanks for all
This is great
In the first question can't two men stand together? Why are there 11 spaces? when there are only 8 people..? I'm sorry it confused me.
If there is no women standing between 2 men then they ARE standing together.
So if if the 3 women are standing in the first 3 spaces, then m3,m4,m5 are standing together.
If
first woman is between m1 and m2, the 2nd woman is between m3 and m4
and last woman is beside m5, then m2 and m3 are standing together
(because there is no woman between them). m4 and m5 are also standing
together for the same reason.
W1 m1 m2 W2 m3 m4 W3 m5
What about above combo?
If there is no women standing between 2 men then they ARE standing together
If my combo is possible then above statement fails.
Think critically bro
thank youuu!!
THANK YOU!!!
thank you!!!
Sorry, I didn't mean it the way you took it. You said nothing wrong. Thanks for your comment and praise.
@AkHarith67 because the men and women are different, they are not treated the same way as letters.
In how many ways can six girls and two boys be arranged in a line if there are at least three girls separating the boys. Help would be much appreciated! Thanks :)
Thank you.
Thank you very much, it helped mi a lot,.
That's okay, thanks for watching
Hi Sir!
I tried to solve first principle by the method taught by you in the last
video. I am getting totally different answer. I rechecked it many a
times.
8 people (5 men+3 women). Total arrangements= 8! = 40320.
Number of arrangements where women are together= 6!*3! = 4320 (Note:
Considering women as a group, we have 6 groups and 3! internal
arrangement for 3 women). Number of arrangements where women are not
together= 40320-4320 = 36,000.
It is quite different from the answer obtained by the method taught
here. i.e. 14,400
I asked the same question.. what is the correct answer..
Hi!
I finally know why your method doesn't work. You only consider one arrangement that is why the answer you found I bigger that the actual value..
Thank you so much the video has been immensely helpful
You're welcome
@@ExamSolutions_Maths I dont understand why you did what you did in the first question? Im really confused and the comments that people make trying to explain it, don't make any sense to me? Im replying to this because I see that this comment was made 1 week ago, so I hope you respond to this as well. Thank you in advance.
Ok... thanks too.
Thank you so much!!
Akhila Acharya That's okay - thank you for looking at it.
thank you so much for this video. now i understand permutation!
Why don't we begin by spacing the three women, which gives us 7 spaces and then fit the men in between?
THX alot! from 2021
I am a little confused as to why in the first question, two or more men cannot stand together? It's not as if mentioned in the question that no two persons of same gender can stand together.
So if we solve according to how I understood the question :
Total permutation of people standing in any order -(permutations of two women insisting on standing together)
5!*3!-(5!*1! ×2-because two women are considered as one unit)
Answer=480
Please tell me if this is correct
It is five years late but I agree to this
good
thanks
@NRGxPHlL You are correct
it helped me a lot..thanks..but why in d first example 5! x 6x5 4 is not 5! x 6! x 5! x 4!??
Permutations is not for you sis
what if no three women are together, so 2 wqomen together is ok
May I ask what they mean by arranging in a row? If 3 women fill any 3spaces, wouldnt there still be spaces between some men? Does this still count as "standing in a row"?
Im kinda confused due to the fact that I've always been learning to draw a fixed amount of spaces based on the total amount of people or items
very well explained, but for the first Q, would we not multiply with 3!, since the women can interchange between themselves as well. Since they are 3 different people. I guess you missed that out by mistake. Or maybe I'm mistaken.
You have mixed in an alternative way to think about this problem: 6choose3 * 3!
The way in the video is such that when one women chooses any of the six vacancies, she can appear in any of those, thus her order as opposed to the other women has been accounted for.
Thanks so much for your videos. They were very helpful. I tried your approach with most questions and got all right except one. The question is: In how many ways can 3girls and 3 boys be seated in a row, if no two girls and no two boys are to occupy adjacent seats?
I did this _ B1 _ B2 _ B3 _ and got 144 (3! * 4 * 3 * 2), but I realized the girls should not have more than 3 available seats for condition to hold. I still did not get the answer right. Please can you help?
Have you got the answer yet?
In the second example, we divided the spaces that can filled by 3 S's by 3!. that i understand. because the 3 S's S1 S2 S3 can be arranged 3! different ways.
But why didnt it happen for the W's in example 1.because cant these W1 W2 W3 also be arranged in 3! different ways. they are different(women), right?
anyone care to explain.
women are distinguishable, letters are not
@MARKNOONAN1 Cheers
Please answer this:
Q. A shelf is to contain 7 different books, of which 4 were written by Booker and 3 by Hardy. Find the number of arrangements in which the first two books at the left-hand endure by the same author.
Please also show me how you did!
Thanks in advance
This may be wrong but here's my answer:
We have -> B B B B H H H
there are 2 possible scenarios:
1. B B _ _ _ _ _
and
2. H H _ _ _ _ _
for 1, the Bs can be arranged 4c2 different ways. the rest can be filled in 5! different ways so we have 4C2*5!
for 2, the Hs can be arranged 3c2 different ways and the rest can be filled in 5! different ways so we have 3c2*5!
add these up to get 1080: 4C2*5! + 3c2*5! = 1080
Please correct me if I'm wrong...
Josh Mozza I understood how you added up the two scenarios but I think the mistake in your solution lies when you are computing the arrangements of the Bs (in scenario 1) and Hs (in scenario 2). Instead of 4C2 and 3C2 respectively shouldn't it be just 2! for both of them. If I am wrong please correct me. Thank you
thank u
Zahidullah Noori 4P2*5! + 3P2*5! = 2160
Why choose 9 places
for the second part, can we write (4! / 2!) * 5C3
Yes you can.
Not 5c3 it would be 5p3
No it will be a permutation as the order matters
Shouldn't you then multiply by 3! for the different ways the women can be arranged?
No. This would be the result if there were only 3 spaces to fill for just 3 women. We have 6 spaces, hence 6x5x4.
For the second one, why is there 9 spaces when there is 7 letters?
I'm wondering the same thing!
As posted above.
You could can do this with 7 spaces but the approach shown in the video is easier.
You would start with all the possible positions for the three S’s where they do not appear together, for which there are 10 possibilities;
_ S _ S _ _ S
S _ S _ S _ _
_ _ S _ S _ S
_ S _ _ S _ S
_ S _ S _ _ S
S _ _ S _ S _
_ S _ S _ S _
S _ S _ _ _ S
S _ _ _ S _ S
S _ _ S _ _ S
The remaining four letter can then be arranged/inserted in the gaps 4!/2! times.
So the total permutations would be 10 x 4!/2! = 120 arrangements, same as in the video.
A PIN-code for a credit card consists of an ordered sequence of 4 digits from
the set {0,1,2,3,4,5,6,7,8,9}. How many such PIN-codes have precisely two
identical digits?
can u help me with this thnx
For question 1: can't some of the men stand together? I'm sure there are enough spaces for the women to not be together right? Why are we forcing women to be in between men? If we do it like how u did it, what would happen to the remaining empty spaces? There's no men or women there but we are still considering that space? Or is there smth I do not understand in ur computation that eliminates those empty spaces for a given order? I feel that the workings of permutation and combination is confusing. In most of the questions we pick people for a fixed amount of space. In the first question, I feel that we are picking spaces for a fixed amount of people. It really confuses me. Please explain how it works. I feel like I'm blindly remembering.
If there is no women standing between 2 men then the 2 men are standing together.
So if if the 3 women are standing in the first 3 spaces, then m3,m4,m5 are standing together.
If first woman is between m1 and m2, the 2nd woman is between m3 and m4 and last woman is beside m5, then m2 and m3 are standing together (because there is no woman between them). m4 and m5 are also standing together for the same reason.
@@tanveerhasan2382 It's surprising how the formula eliminates the empty spaces. It's hard to picture what it means to multiply two permutations.
If the first place is sat in by a boy B1 _ B2 _ B3 _ Try 3! for the number of ways the boys can arrange themselves then the girls can arrange themselves 3! ways, making 3! x 3! = 36 ways. But if a girl sits in the first place G1 _ G2 _ G3 _ then the same argument is repeated again. So it should now be 3! x3!x2 = 72. Hope that agrees with your answer.
Why there is x2 in the second answer?
Watching this after 12 year