These lectures are brilliant! I have hardly a week left before my exams, and I never really managed to understand permutations all year. Until I saw these lectures. Hats off to you sir, for such excellent explanations!
Im taking CIE A levels and this is the hardest topic for me. This chapter is in AS but I think that it's the hardest of both AS and A2. Maybe you won't reply to me, but I have to say, thank you for making videos on Perm and Comb. I can't find any other videos that gives a full lesson on the Perm and Comb. of my syllabus's standard.
@varietychannel5000 These examples are the easiest examples that can be provided for our Permutation and Combination questions. U should take a look at our statistics papers. There will be permutation and combination questions in all the papers
if the two L aren't rearranged in question 1 because you don't notice the difference between them how come in the word parallel you times by 2 and 3 factorial if there is no difference in order of L's and A's
Because the two L's in HELLO are the same regardless of their arrangement (LL and LL are the same). In the last example, the arrangements R and P can be either RP or PR thus they can be arranged 2! ways.
For parallel, why didnt you multiply the end result by 7 because the P and R could by together after one letter, or two letters, like : A(PR)ALLLE and AA(PR)LLLE etc.
hi, why didnt u divide 5!X3! by 4! for the no of repeated men. in people questions do we not have to do that? and if it were all different letters in the last example, wud we have to divide again? i have a stat exam tomorrow!
But u alr count them as one by doing 4! instead of 5!. If u do 5! It means that u considered their arrangements. L1 L2 or L2 L1. So u have to divide 2 to get L1L2 only or L L if u like.
@@oneinabillion654 if you didnt consider them as the same objects, then you would have to divide by two, but because the restriction makes them one item, we neither multiply by two or divide
For some, there can be repetition of letters but for other u can't. Like my CIE A level statistics, we cannot have repetition so we do this: (4!2!)/2! Which cancels to 4!. Again, this is when there is no repetition of letters
the problem in which the items are not together..i tried to use that method in a,b,c,d where a and b must not be together...i got wrong answer. i also tried to get 8! - (no. of ways in which a and b sit together)= the no. of ways they will not be together...i didnt get the same answer
You have to divide when you see identical letters. when some of the letters are repeating themselves, you divide by their total count factorial. It's vice versa for the 2nd and any other similar example.
Bro they are literally the same, It can only Be possible when Two letters are different like The last question (PR and RP) Whats the point of (LL and LL) haha
I think the first example is wrong. I can prove it. Lets take a simple example instead. We have *WOW*. How many different arrangements can we make? 1)We use 3!÷2! we got 3. 2)We consider W as 1 we have 2! and we got 2 So which one is correct? Lets arrange them to find out which working is correct. *WOW* can be arranged into: OWW WOW WWO Theres 3 arrangements! So we use the first working! I hope this helps
My solution is correct. Your workings for the specific situations are correct but you have missed the point of the video. The video is demonstrating how many ways the two letters L stay together, NOT how many different arrangements of the letters in HELLO which would be 5!/2! (Your case 1 for WOW). However I required the LL in my example to be together and so it is 4! as described in the video which leads to 24. This is the case you demonstrated in your example 2. I hope that clears this up for you. Please let me know and I will then remove the comment as it may cause confusion for others if they think my solution is wrong. Thank you.
May God bless you seriously man thank you so much I needed this so badly I have a test this Wednesday and I hope this helps (I believe it really did) you explain it way easier than my teacher does idk what she even talks about. Thank you tho
Why does he counts the letter L as 1 if there are 2? In combinations, two identical items count as 1, but in this example we are revising permutations examples. Some can clarify this example. Please
A machine is used to generate codes consisting of 5 letters (A, B, C, D, E), 3 digits (6, 7, 8) and 2 special characters (#, @). Find the total number of codes that can be generated i. if the letters, digits and special characters must be grouped together
On the HELLO example, shouldn't you multiply the answer by 2 coz the LL can be arranged in two different ways as we had seen in the past tutorials? That means the answer should be 48.... Is that right? please advice
No because we learn that when they are identical u have to divide it and if you multiplied by 2 you would also have to divide the 2 by 2! Again which is 1
Hi! I’d like to answer your question. We can listen again to 6:35. He explained that the A’s could rearrange themselves 2! ways and the L’s could rearrange themselves 3! times. Hope this helps.
No as the L's would still look the same if arranged amongst themselves. What it the difference between HELOL and HELOL none you can notice but your argument has factored in that they are different. So do not multiply by 2!. I hope that makes sense.
Suppose if we had the letter REMEMBRANCE and it asked for arrangements in which all the vowels( EEAE) were next to each other. Would it be (8! x 4!)/2!2!?
@@ExamSolutions_Maths I've noticed that the vowels EEAE contain 3 Es which would rearrange within themselves. So is it (8! x 4!/3!) / 2! 2! ? Since 4!/3! gives us a 4.
When I arranged the consonants together I got "(C L L Q M) O O U U I" so what I did is 6!/2!*2! for the word as a whole then I multiplied it by 5!/2! for the letters in the bracket which have to be together so I got as a whole " (6!*5!)/(2!*2!*2!) and so the answer I got is 10800 ways. I hope it is correct xD
I have a question, in the 1st and 2nd questions surely the other letters/ the men can arrange themselves in different orders around the LL/the women as well, so how come that is not accounted for?
Hello Professor, why you haven't consider the 4 men as 1 ensemble like you did with the women so instead of saying 5! we say 2! then because men and women could permutate among others then we better say 2!3!4!. Please advise.
Thank you for the explanation but am still a little confused regarding the first and last example ...HELLO making sure that LL are always together in the arrangement we have 4!of arranging the word Hello but why didn't we multiple it by 2 since there are two way arranging the letter L just like how we did in the last example PARALLEL were we treated the La to be different even though the are same ..
Hi Mimi, that's because swapping the LL wouldn't make a difference as it looks the same. Maybe you could swap the Ls if you label it differently like L1 and L2. Always good to check answer sheets of past papers if you can find a similar question. See how they mark it.
No dude we considered the 3 women as only one woman so we are now arranging 5 people 4 of them are men but at the end we multiplied by 3 factorial as it is the number of ways the three women can arrange themselves between each other
That would be the case if there were no restrictions. It would have been 5!/2! . This is NOT that case. The L's must stick together. Try listing all the arrangements and you will see what I mean.
These lectures are brilliant! I have hardly a week left before my exams, and I never really managed to understand permutations all year. Until I saw these lectures. Hats off to you sir, for such excellent explanations!
Same here!
that is because same alphabets are together so no need to rearrange them!
I''m having an "aha" moment. Thanks for your help.
Im taking CIE A levels and this is the hardest topic for me. This chapter is in AS but I think that it's the hardest of both AS and A2. Maybe you won't reply to me, but I have to say, thank you for making videos on Perm and Comb. I can't find any other videos that gives a full lesson on the Perm and Comb. of my syllabus's standard.
It’s easy u just need to focus and practice
@varietychannel5000 These examples are the easiest examples that can be provided for our Permutation and Combination questions. U should take a look at our statistics papers. There will be permutation and combination questions in all the papers
Not really the hardest, but definitely the trickiest.
They could but would you notice the difference? No. LL is the same as LL even if I swapped the L's around, so don't count it.
Ohh ok…would our teachers also count it this same way coz I don’t want to fail the exam
if the two L aren't rearranged in question 1 because you don't notice the difference between them how come in the word parallel you times by 2 and 3 factorial if there is no difference in order of L's and A's
please answer???
The 2 L's in hello are treated like one letter, whereas in PARALLEL they are separate letters.
my g
but in hello we also have got 2 L's y u r not considering them as L1 and L2 as u did in the case of women in.second example
stfu vegan
i was also thinking the same
U have to think logically on this. Every woman are different in the real world but L's are always same
i'm wondering why he didn't multiply the answer by 2! and im confused
@@markom3954 stupid as*hole
@killthrillabs Thanks
Because the two L's in HELLO are the same regardless of their arrangement (LL and LL are the same). In the last example, the arrangements R and P can be either RP or PR thus they can be arranged 2! ways.
Yes. I have the same doubt
Why dont we divide by 2! Since they are repeated letters
For parallel, why didnt you multiply the end result by 7 because the P and R could by together after one letter, or two letters, like :
A(PR)ALLLE and AA(PR)LLLE etc.
hi, why didnt u divide 5!X3! by 4! for the no of repeated men. in people questions do we not have to do that? and if it were all different letters in the last example, wud we have to divide again? i have a stat exam tomorrow!
Why didn't we divide the first example of "H E L L O" by 2! because the L's are same.
Because the Ls' rearrangements are 2! And L repeats twice 2! so they cancel out because you are multiplying 4! with 2! and then dividing by 2!
+Yasir Siddiqui i think its because the LL is 1 item
But u alr count them as one by doing 4! instead of 5!. If u do 5! It means that u considered their arrangements. L1 L2 or L2 L1. So u have to divide 2 to get L1L2 only or L L if u like.
@@oneinabillion654 if you didnt consider them as the same objects, then you would have to divide by two, but because the restriction makes them one item, we neither multiply by two or divide
TRΔXX TRΔXX why it was not done like 5!/2!.
Because it is done so in the last example of PRALLEL
@fadedglory94 No because the L's are the same.
Sir why we do not divide the two L's by 2!?
That's what I like to hear!
Could you use the nPr notation for these examples and if so how?
7!=7P7
@fadedglory94 That's right
@ExamSolutions how to do it for PARALLEL if I want no TWO L be together ?
Why in the last ques 3! And 2! Are divided? Can anyone explain
Two L's in 1st example are different L's. So they can replace each other and the answer is 4!*2!=48. Am I right?
YA MAN...THATS WHAT I'M ARGU'NG
Roman Fesenko yes you're right
It depends on which exam u are studying for
For some, there can be repetition of letters but for other u can't. Like my CIE A level statistics, we cannot have repetition so we do this:
(4!2!)/2! Which cancels to 4!. Again, this is when there is no repetition of letters
For all questions on this video, there is no repetition. So keep that in mind
the problem in which the items are not together..i tried to use that method in a,b,c,d where a and b must not be together...i got wrong answer. i also tried to get 8! - (no. of ways in which a and b sit together)= the no. of ways they will not be together...i didnt get the same answer
In how many ways can six cats and five dogs stand in a line so that all the cat stand side by side and all the dogs stand side by side?
thanks for the tutorial. so how do we know when to divide by the various arrangements like the last example and when not to like the 2nd example
You have to divide when you see identical letters. when some of the letters are repeating themselves, you divide by their total count factorial. It's vice versa for the 2nd and any other similar example.
but on your first example on HELLO shouldn't we also multiply with 2! for the L's
thanks
Thank you! It really helped.
There 6red balls and 3 green balls how many ways can they be arranged if the green ball should be together
If they are two l's does that mean they can be arranged 2! Factorial ways?
Wow.This was amazing ,Thank You so much
very good and nice explaining method that you described in the video. many many thanks
Thank you
You Sir, you just saved my exam :') Thanx for your big help. Cheers :)
Why dont we divide hello by 2! Since the ls are repeated
The first question, 48 should be the ans. The two Ls can be rearranged in each case!
Bro they are literally the same, It can only Be possible when Two letters are different like The last question (PR and RP)
Whats the point of (LL and LL) haha
Confusing. I need to re-watch!
4! Right? But why the answer is 24? It must be 48
I think the first example is wrong. I can prove it.
Lets take a simple example instead. We have *WOW*. How many different arrangements can we make?
1)We use 3!÷2! we got 3.
2)We consider W as 1 we have 2! and we got 2
So which one is correct? Lets arrange them to find out which working is correct.
*WOW* can be arranged into:
OWW
WOW
WWO
Theres 3 arrangements! So we use the first working!
I hope this helps
My solution is correct. Your workings for the specific situations are correct but you have missed the point of the video. The video is demonstrating how many ways the two letters L stay together, NOT how many different arrangements of the letters in HELLO which would be 5!/2! (Your case 1 for WOW). However I required the LL in my example to be together and so it is 4! as described in the video which leads to 24. This is the case you demonstrated in your example 2. I hope that clears this up for you. Please let me know and I will then remove the comment as it may cause confusion for others if they think my solution is wrong. Thank you.
May God bless you seriously man thank you so much I needed this so badly I have a test this Wednesday and I hope this helps (I believe it really did) you explain it way easier than my teacher does idk what she even talks about. Thank you tho
Best wishes for Wednesday
thaaanks
Why does he counts the letter L as 1 if there are 2?
In combinations, two identical items count as 1, but in this example we are revising permutations examples. Some can clarify this example. Please
In permutation questions, when you come across identical objects, we consider them as one. The 1 denotes that they are identical or same.
Appreciate the colour coding
Whenever I view and watch your videos, I know I will get what I need (They are productive). For that, I thank you.
A machine is used to generate codes consisting of 5 letters (A, B, C, D, E),
3 digits (6, 7, 8) and 2 special characters (#, @).
Find the total number of codes that can be generated
i. if the letters, digits and special characters must be grouped together
thank you sir
for such a wonderful explanation
You're welcome. Thanks for watching.
very clear!!
Why is it that the first question on "HELLO", you never divided the answer by 2! as we have 2Ls?
vry helpful m8 thx!
thanks for listening.. yeah right thanks for teaching with out you videos i would definitely fail
Thank you for saving my life! You deserve to be in heaven
because you treat the women as one person, and the men can sit anywhere around them. hence not needing to divide by 4!
@ExamSolutions so if the letters that needed to be together are the same, we do not need to multiply it by 2! ?
What if the questions says that P and R cant be together?
why does the HELLO words doesn't need to be multiplied by 2! ? isn't it the same as the last one?
Helpful, thanks!
Cheers
thank you! it really helps! I'm done with my tutorial!
love u
Respect
On the HELLO example, shouldn't you multiply the answer by 2 coz the LL can be arranged in two different ways as we had seen in the past tutorials? That means the answer should be 48.... Is that right? please advice
No because we learn that when they are identical u have to divide it and if you multiplied by 2 you would also have to divide the 2 by 2! Again which is 1
Really the best explanation I have seen sincere thanks
Very clear and concise thank you.
Why the word ‘hello’, you didnt divide 2! while in the word ‘parallel’ you divided it to 2! & 3! ?
Hi! I’d like to answer your question. We can listen again to 6:35. He explained that the A’s could rearrange themselves 2! ways and the L’s could rearrange themselves 3! times. Hope this helps.
My concepts just got clearer.
THANK YOU SO MUCH!!!!!!!!
Thank you so much
@ExamSolutions same question from my side too.
for the HELLO one,
wouldn't it be 2! (arrangements within LL themselves as they're interchangeable) x 4! ?
No as the L's would still look the same if arranged amongst themselves. What it the difference between HELOL and HELOL none you can notice but your argument has factored in that they are different. So do not multiply by 2!. I hope that makes sense.
@@ExamSolutions_Maths ah I see, thank you!
Why didn't you include a 4! For the number of ways men could permute amongst themselves?
That was counted in when I did 5!
@@ExamSolutions_Maths thank you soo much. 🤗. Your solution did work like a charm!
Thanks a lot
You saved us
Hello! I have a question. In example 1, why is it 4! and not 4!x2! like that of example 2? Hope to read your response. Thank you!
Hello :D This is because, in the 1st example, the restricted items were the same (L's) and not different from each other.
@@ExamSolutions_MathsI still don’t understand…
It should be 4!*2!
Thanks!
Suppose if we had the letter REMEMBRANCE and it asked for arrangements in which all the vowels( EEAE) were next to each other.
Would it be (8! x 4!)/2!2!?
I disagree. I make it (8! x 4)/ 2! 2! as when the EEAE are together they can be arranged only in 4 ways not 4! ways
@@ExamSolutions_Maths I've noticed that the vowels EEAE contain 3 Es which would rearrange within themselves.
So is it (8! x 4!/3!) / 2! 2! ?
Since 4!/3! gives us a 4.
@@s1nister688 That's how I got the 4.
@@ExamSolutions_Maths oh I see. Thanks a lot.
@ExamSolutions thanks!
in the second example
why multiply 5! by 3! and not divide by 3! ?
because the women are indistinguishable. They can be rearranged in 3! ways.
Great revision, thanks =].
how many ways the letter COLLOQUIUM can be arranged if the consonants have to be next to each other.
When I arranged the consonants together I got "(C L L Q M) O O U U I" so what I did is 6!/2!*2! for the word as a whole then I multiplied it by 5!/2! for the letters in the bracket which have to be together so I got as a whole " (6!*5!)/(2!*2!*2!) and so the answer I got is 10800 ways. I hope it is correct xD
CLLQM OOUIU => (6! x 5! x 2!) / (2! x 2!)
I think it is
6!/(2!*2!)*5!/2!
Dexter Ns yes I think so too
i think,there should be 6!5!/2!2!2! ways or arrangements
It helps me to answer problems that my math teacher gives me... Thank you😊❤
The first answer is wrong
No it wasn't
I have a question, in the 1st and 2nd questions surely the other letters/ the men can arrange themselves in different orders around the LL/the women as well, so how come that is not accounted for?
Hello Professor, why you haven't consider the 4 men as 1 ensemble like you did with the women so instead of saying 5! we say 2! then because men and women could permutate among others then we better say 2!3!4!. Please advise.
Tamer Aziz there are different men. He just counted women as one because they have to be with each other anywhere in the arrangement
Thank you for the explanation but am still a little confused regarding the first and last example ...HELLO making sure that LL are always together in the arrangement we have 4!of arranging the word Hello but why didn't we multiple it by 2 since there are two way arranging the letter L just like how we did in the last example PARALLEL were we treated the La to be different even though the are same ..
Hi Mimi, that's because swapping the LL wouldn't make a difference as it looks the same. Maybe you could swap the Ls if you label it differently like L1 and L2. Always good to check answer sheets of past papers if you can find a similar question. See how they mark it.
In the 2nd example, shouldnt it be 5!x3!x4!? Because there are 4! ways to rearrange the men, which was not considered?
No dude we considered the 3 women as only one woman so we are now arranging 5 people 4 of them are men but at the end we multiplied by 3 factorial as it is the number of ways the three women can arrange themselves between each other
Why you didnt divide HELLO with 2!?,bacuse L is repeating and it looks same.
That would be the case if there were no restrictions. It would have been 5!/2! . This is NOT that case. The L's must stick together. Try listing all the arrangements and you will see what I mean.