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
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
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.
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.
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!
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
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
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
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
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
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
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
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.
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
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.
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.
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
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.
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.
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.
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?
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
