Combinations - Further methods : ExamSolutions
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- Опубликовано: 1 окт 2024
- Tutorial on further methods of combinations.
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I have watched so many of your videos. It is like a revision for me. You are so structured, and your words map directly to definitions or concepts. . It has helped me understand clearly & not forget. I really enjoy your teaching. Your students are the lucky to have you as teacher. Once again, thank you.
+Good DeedsLeadTo Thank you for your comments and I am pleased to hear that the videos are helping. Just for the record, I am not an employed teacher though. You are my students.
+ExamSolutions 74,000 students, thats a lot :P
It's a bit like probability. When using the and statement multiply and when using or add.
Not at all. Having 0 A's is a possibility and so we cannot rule it out from the possible selections of 5 letters.
Much the same. Break it down in all the different stages and add them. Far too difficult to explain it here though. Good idea to do another video on this at some stage.
No. It was a coincidence that 8C3 gave the same value.
I dreadfully love this accent.very lovely, factorial😁😁
I didn't. I considered the possibilities of no A's, 1A, 2A's etc.
sir,i dont get one thing about the advantage question. In the case when there is 1 or 2 A's,when you divide by X!, X being the number of spaces left,why dont you include the space taken by the A in the start? Eg. what i thought when i tried the question,for 1 A,i did (6 x 5 x 4 x 3) / (5!) because there are 5 spaces taken by the 5 letters which can be arranged in any way...isnt that right..? please reply asap...
Oh that make sense. THANKS YOU SO MUCH FOR REPLYING AND REPLYING SO QUICKLY
I enjoyed your videos so much, I followed step by step of your teachings and I found permutations easy for me now. And I wish you could make another video on geometry proofs.
Thank you.
how do you know when to times or add ?
I wonder why the following wouldn't work for the last problem:
The No of permutations is 9P5/3! (since there are 3 As). Therefore the number of combinations must be 9P5/(3!*5!) since there are 5 'places'. This gives 21, quite far from 56. Any insight would be appreciated :)
this 21 answer is for selection of all distinct letters.
Why don't you multiply them together instead of adding them up? I am a bit confused
Because they are not separate conditions but rather a fused one. Notice the AND instead of something else.
hello! i just wanna ask yk for the question with the word 'advantage' right...I wanna ask what happens if there is more than 1 letter repeating, like MISSISSIPPI for example???
Why weren't you using the choose formula for the As. Isn't it for.when we say 6c4 we also have to say 3c1 to make up the 5 letters
I wanted to ask that in the second example, as we add an A, the number of factorial arrangments they can form amongst themseleves is decreasing, I am unable to understand the relation, Can anyone please explain/refer to the a video in the playlist that I should re-visit?
I believe the answer must be 126 ways in second example.. as the A's can be placed anywhere in the 5 places.. kindly advise..
Hello Sir, Thank you so much for your help. But I am a bit confused on the second example. Why do we divide by 3! on where there are 2As, 4! on 1A etc. why not 5! as...
...the A can also be rearranged. Btw thank you so much for making these videos, it has really helped me and as I can see a lot more people as well.
6C3 x 6C1 = 20 x 6 = 120
9:25, i get confused when to add and when to multiply in the end .
How to Solve the Question like SUCCESS ...???? that need to be selected?????
plz reply as soon as possible and yeah thanx for this one...
Im confused for the first question why do you times the two binomials, yet for the 2nd question you add them all up ?
I didnt understand why u said we dont need arrangements and he put only 6×5×4×3×2 in 0A but not 1 pls explaim
about the second part, how would you solve it if there are two, three or four and so on different sets of repeating units? for example, if 4 letters are to be chosen from 'SEQUENCES' how many combinations are there?
How come sometimes you have to divide the repetition (like in the example of letters) while sometimes you dont have to divide it when dealing questions like women are selecting to be together? (Arent they two seperate women afterall?
There are 6 woman and 6 men in a club. Four members have to be chosen for a committee.
how many consist of 3women and 1man?
Thoughts?
..yu make maths look so easy..hats off
Your videos are amazing i would like to know if there are words that appear most of the time that help to differentiate weather the question requires your knowledge on combination or permutation and if they are i would like to know those words under the cie As and A level Statistics Thank you very much.
How would you go about doing the number of permutations of selecting 5 letters from 'advantage', rather than combinations?
what about the fact that the 2 and 3 A's can also rearrange among themselves
Order doesn't matter in combinations
how can we do by subtracting by A 4 and A 5 from total combination..?
nice video can you make some 3 variables linear equations word problems thank you
Thank you for a crystal clear explanation. I was struggling to find out which step is "C" hiding when I turned to solution. Thank you once again Sir.
So we always find the repeating number in the second question?
Thanxx......but what if more than one letter repeats.??
you broke it up into parts depending on the number of A's in the word, and then put the A in the "first" letter slot, surely there is a possibility that the A could be in any of the other slots? meaning you would have to multiply by a further 5C1
Not at all. It is a combinations problem and in any group of A's I considered, irrespective of the position of the A's there would be one combination of these. It is the remaining letters that matter. Try and write out all the combinations say for the 3 A's and you will most probably see more clearly the point I am trying to make.
I thought that the mean of 20 would be (20+1 /2)= 10.5 th term not the 10th term. Can you explain why you put the 10th? :)
I am unsure what you mean, sorry and what that has to do with this video.
That was suppose to be a permutation question, my apology.
Six men and three women are standing in a queue.
Three of the people in a queue are chosen to take part in a survey. How many different choices are there if at least one woman must be included?
Sir please make a video tutorial on this question.
If you want an answer, please post in facebook.com/groups/566605273474018/
Very helpful 👌
sir...thank you so much.
thx for ur efforts towards our bright future n God bless u
very very helpful
thanks
that helps a lot thank you
Could somebody explain to me why we can't apply the nCr formula but have to consider each event separately? Why would the formula fail to capture the events: 1) one A, 2) two As, 3) three As ??
It's because if we didn't treat them separately we would be counting all the As as distinct and as a result would count a bunch of non-distinct selections. E.g. A1 A2 D V T would be counted as distinct from A2 A1 D V T.
Why are we ignoring the A???
I've mastered everything in maths, both pure and stats, except for this stupid combination
If we didn't consider the As separately it would be as if we treated them all as distinct letters. E.g. the combo A1,D,V,N,T would be counted separately from A2, D,V,N,T . Get it?
@@Unstable_Diffusion89 I think I kinda get what u mean
In your second example, is there no way of directly getting the answer instead of splitting the question into 4 parts?
+Yash Verma Not as far as I know.