Omg, the relationships are totally blowing my mind. I am rediscovering math with Eddie Woo as my guide and I can see the path so much clearer now. Thank you-WOO! :D
Loving how confident and vibe when he taught his student. Also, his lecture is really logical and detailed though you felt like you got caught up in the his story,
In an attempt to explain the last part as I was a bit confused too because haven't touched the binomial theorem before permutations/ arrangements. Look up (a+b)^6, that is the equation the KKKLLL(aaabbb) "name" relates to. There is an part of the result of the equation that gives a^3b^3, meaning aaabbb(KKKLLL), where 6 choose 3 "b"'s, where the amount of "b"'s tell us how many items to choose from the lot or 6 permute 6 of the lot gives us 20, the coefficient of the part of the result, and there is the corelation. I haven't looked too much up about this, this is the best I could come up with so I hope it's enough. 😁 You should watch the beggining of his Binomial Theorem playlist atleast for more insight.
I am not a maths master and he is awesome and I know that out there they use calculators for calculation which is good but i just wanted to tell that if u asked what's 5! Or 6! Like that in India or even 720 /4 u will be judged as a teacher. I don't say he is bad a teacher good at concepts is far better than a teacher just knowing everything himself or just knowing the numbers... I repeat this teacher is great
Picking up from, 6!/[3!(6-3)!] = 6C3. While mathematically correct, the equality is perhaps a little misleading. That being said, 6!/[3!(6-3)!] = 20, which implies that there are 20 distinct permutation of the letters KKKLLL (if we select 6 letters). If we now select 6 letters from the 6 letters (KKKLLL) we will have 6C6 = 1 combination. So, my point is, do not confuse permutations with combinations.
No, the equality is correct, because the question is asking for combinations, not permutations. It's 6C3 because the Ls can only go in the blanks left by the Ks. If it were KKKLLLL, the answer would be 7C3=7C4 (both give the same result) for the same reason.
Thanks fort the Video Mr. Woo! I just had a question: suppose from the word KELLY, I picked only two letters e.g KE,KL, etc. Would you approach repetition the same way? Thanks in advance!
I think there is a mistake at 6:57, i mean Ls and Ys can not switch together either it will be a different word so the number of ways the one word can repeat is how many ways can ls switch together plus how many ways can ys switch together not times, it just happened that s factorial plus 2 factorial is equal to two factorial multiplied by two factorial
Hello Mr.Woo I have a question regarding the last part about the 2 letter thing so im given 6 letters and im asked to to make from it a 2-letter word so this is 6!/(6-2)! and then i have repetitions .... im kinda lost can you clarify the question of the 2 letter .. i just want to know what was the question and thank you in advance for your intuitive videos
At 8:00 how are u treating words like expressions in bionomial expansions, i mean isn't KKLKLL different from KKKLLL as a word but both are the same as bionomail expansion terms K^3B^3??
Sir please help me with this. In how many different ways can five boys and five girls sit around a table such that the boys and girls are at alternate positions??
@@gameswithtags802 that answer is off by 2778 assuming the question states the table is circular 🙂 with a circular table, we can fix one person in place because the rotation of the table doesn't matter, leaving 9 slots to fill. If a boy was fixed in place then there are 4 possible spots to place the remaining 4 boys, which is 4! ways. Then put the 5 girls in the 5 remaining spots, which is 5! ways. 1x4!x5! = 2800 (total ways)
I still didn't understand this sir,can anyone please explain me in the comments PLEASE,BEGGING YOU because I am currently breaking down cuz I am not able to umderstand anything
Take the eg.Kelly, here, out of 5 letters i need to arrange 5 letter, therefore i can arrange them in 5P5 ways. But also there are two identical letters (i.e. L and L). So even if we arrange them in different ways they word is same. Those L's can arrange themselves in 2P2 ways = 2! ways = 2 ways. (1st L, 2nd L or 2nd L, 1st). which is the same. So we divide 5P5 by 2! to get number of ways to arrange the letters.
This guy is the greatest math teacher on earth
Ikr Way better then mine for sure
@@ritamkarmakar7796 r
yeah I'm agree😊
So so so true❤
I am 44 years old and this is the first time I've understood how to work on problems if there are repeating letters
Omg, the relationships are totally blowing my mind. I am rediscovering math with Eddie Woo as my guide and I can see the path so much clearer now. Thank you-WOO! :D
Such a brilliant lecture! Now I am more confident about the exam tomorrow. Big thanks for you, Eddie Woo!!
Loving how confident and vibe when he taught his student. Also, his lecture is really logical and detailed though you felt like you got caught up in the his story,
Thank you Eddie! You are the best teacher on RUclips
You are great! Really catch the attention and make it interesting.
another great lesson, especially the way you explained over counting when there is repetition
thanks so much sir! really hate this topic but you've made it just that wee bit better :)
Raymond Zhu
You mean just that woo better ahaha
Kellly
For one permutation there are extra 5 permutations .
Why don't we eliminate 5 instead of 6.
@@curryrice9577 hahaha
Kelly might never forget anything from this lesson like me
Best math teacher ever
Excellent explanation, got it crystal clear in my mind now
Great teacher you are!! Thank You!
In an attempt to explain the last part as I was a bit confused too because haven't touched the binomial theorem before permutations/ arrangements. Look up (a+b)^6, that is the equation the KKKLLL(aaabbb) "name" relates to. There is an part of the result of the equation that gives a^3b^3, meaning aaabbb(KKKLLL), where 6 choose 3 "b"'s, where the amount of "b"'s tell us how many items to choose from the lot or 6 permute 6 of the lot gives us 20, the coefficient of the part of the result, and there is the corelation. I haven't looked too much up about this, this is the best I could come up with so I hope it's enough. 😁 You should watch the beggining of his Binomial Theorem playlist atleast for more insight.
Thank you for this! I haven't touched it either (though, I'm sure you have by now 😅) and this made me understand the last bit more!
Perfectly explained Sir👌
Excellent!
Sir you're the most handsome maths teachers I've ever seen and you're the best maths teacher also.
I am not a maths master and he is awesome and I know that out there they use calculators for calculation which is good but i just wanted to tell that if u asked what's 5! Or 6! Like that in India or even 720 /4 u will be judged as a teacher. I don't say he is bad a teacher good at concepts is far better than a teacher just knowing everything himself or just knowing the numbers... I repeat this teacher is great
I think he was asking the class so that he knew they were paying attention and doing the steps along with him
3:34
Me nodding yes enthusiastically🤓
Picking up from, 6!/[3!(6-3)!] = 6C3. While mathematically correct, the equality is perhaps a little misleading. That being said, 6!/[3!(6-3)!] = 20, which implies that there are 20 distinct permutation of the letters KKKLLL (if we select 6 letters). If we now select 6 letters from the 6 letters (KKKLLL) we will have 6C6 = 1 combination. So, my point is, do not confuse permutations with combinations.
No, the equality is correct, because the question is asking for combinations, not permutations. It's 6C3 because the Ls can only go in the blanks left by the Ks. If it were KKKLLLL, the answer would be 7C3=7C4 (both give the same result) for the same reason.
I never knew that binomial distribution gives a hint about the probability
Thanks fort the Video Mr. Woo! I just had a question: suppose from the word KELLY, I picked only two letters e.g KE,KL, etc. Would you approach repetition the same way? Thanks in advance!
Beautiful explanation of (a+b)^3
The only video with 0 dislikes.
Just realized that
Great!
I think there is a mistake at 6:57, i mean Ls and Ys can not switch together either it will be a different word so the number of ways the one word can repeat is how many ways can ls switch together plus how many ways can ys switch together not times, it just happened that s factorial plus 2 factorial is equal to two factorial multiplied by two factorial
legend
i wish you were my teacher 💕
thanks sir!
Hello Mr.Woo
I have a question regarding the last part about the 2 letter thing
so im given 6 letters and im asked to to make from it a 2-letter word
so this is 6!/(6-2)! and then i have repetitions .... im kinda lost
can you clarify the question of the 2 letter .. i just want to know what was the question
and thank you in advance for your intuitive videos
Shit I ended up learning binomial theorem properly too with this video damn
At 8:00 how are u treating words like expressions in bionomial expansions, i mean isn't KKLKLL different from KKKLLL as a word but both are the same as bionomail expansion terms K^3B^3??
Kellly
For one permutation there are extra 5 permutations .
Why don't we eliminate 5 instead of 6.
You do eliminate 5. If the denominator is 6 than its like you multiply it with 1/6. In other words, you take 1 out of 6.
why am i not able to apply combinaton on kellyy? can anyone answer
ikr
What does he mean when he says 'overcounted'?
nvm...i think i get it. When he says he overcounted, he's referring to the identical permutations that were included
@@matthewsmith1571 Exactly he means that we have counted more than they are supposed to be
Sir please help me with this.
In how many different ways can five boys and five girls sit around a table such that the boys and girls are at alternate positions??
only 2 ways???
@@gameswithtags802 that answer is off by 2778 assuming the question states the table is circular 🙂
with a circular table, we can fix one person in place because the rotation of the table doesn't matter, leaving 9 slots to fill.
If a boy was fixed in place then there are 4 possible spots to place the remaining 4 boys, which is 4! ways. Then put the 5 girls in the 5 remaining spots, which is 5! ways.
1x4!x5! = 2800 (total ways)
That is Variation and not Permutation!
I still didn't understand this sir,can anyone please explain me in the comments
PLEASE,BEGGING YOU because I am currently breaking down cuz I am not able to umderstand anything
Take the eg.Kelly, here, out of 5 letters i need to arrange 5 letter, therefore i can arrange them in 5P5 ways. But also there are two identical letters (i.e. L and L). So even if we arrange them in different ways they word is same. Those L's can arrange themselves in 2P2 ways = 2! ways = 2 ways. (1st L, 2nd L or 2nd L, 1st). which is the same. So we divide 5P5 by 2! to get number of ways to arrange the letters.
Hope you understood. And see the vid again