Here's how I learnt it from the video and processed it. When there are lesser positions than the elements to pick or choose from we can’t simply divide by the overcounting. This is because it is not certain that for each arrangement there is the repeated element. For ex.: ALGEBRA and pick/choose 3. But when we do 7C3/2! and 7P3/2!, we get a fractional number and that does not make sense as there can either be a possibility or none, not in between. So to solve it we divide it into sub-arrangements and then add all them up. For Combination, First, no As: 5 total letters and 3 positions so 5C3 = 10 Then, 1 A: This A takes a guaranteed slot so only 2 positions left, 5C2 = 10 Then, 2 A: Similarly to 1 A, 5C1 = 5 And total is 25, which is our answer. As we see different types of combinations, ones without A, ones with a single A and ones with one more A as that’s the total no. of As, and adding them up are the total combinations for 3 positions. Alternatively, we could directly just take 1 A and say ALGEBR, and in that case we just simply do 6C3, and this includes combinations with 1 A and no A. Which can be seen as it equals 20. Now we only need to see combinations with both As and when 2 A take up the slots, we are left with 1 position and 5 letters giving us 5 and total being 25. Similarly, for Permutations, First no As, 5P3 = 60 Then 1 A, 5P2 = 20 However, since order matters here, the 1 A’s place matters as well, as we know A _ _ _ A _ _ _ A are the 3 positions, we multiply 5P2 with 3 Then 2 A, 5P1 = 5 Similarly, these 2 A’s non identical positons matter as well and that gives us 15. Total is 135. Alternatively, we can directly take ALGEBR and that gives us 6P3 which is the sum of permutations achieved through 1 A and no As with order in mind, and it is indeed equal to 120 which we got from 5P3 + 5P2*3. And we follow the same last step which gives us 15 and the total is 135.
@@tokyo7671 That's because 2 of the A are repeating, as explained in the video and my comment, if we simply do 7P3 we are overcounting. We know the 2 A in ALGEBRA are identical, and we need just 3 letters, so if we do AAR or AAR, it is the same thing, and we have overcounted by 2. Normally it wouldn't be a problem if we only had 3 letters and 3 positions, we can just divide by 2 leaving only 1 AAR as the only permutations would be AAR AAR ARA ARA RAA RAA and dividing by 2 removes half of them, which works because there are 2 duplicates for all elements. We aren't really dividing by 2, instead we are *not* counting the duplicates, which is a different thing because not counting means we don't include them in our permutations but division means removing them after they are generated. We can divide the final permutation count if we overcount by an exact figure, like 2 or 3 or so on and that works, that is to say division is a shortcut of reaching the same thing but it is merely a shortcut, not the actual method. But we have 7 elements here, and now if we say 7 Pick 3, we have overcounted, but by an unknown amount, as if we simply divide it by 2 we will remove permutations that don't include 2 identical A. Like EBR or LRB etc., and simply halving them isn't guaranteed to take only 1 AAR of 2 AAR. Assume the only permutations were LBR ERB AAR AAR GEB RBA then if you simply divide them by 2 you are saying you didn't count LBR AAR AAR, but as you can see, it always includes a permutation which shouldn't be removed, it is not an overcount. Then what do we divide the 6 permutations in the example to only take 1 AAR out of the 6 elements ? The answer is 1/6 which is a fraction, we can't really define a fraction, division was just a shortcut of saying we didn't generate x permutations and when it is a fraction, it is incorrect as you can either generate 1 permutation or none, but not 1.5 permutations. So to solve this problem, we build up from a case where we can't overcount. We find all the permutations with 3 elements, then 4 ,then 5, and then 6. We start at 3 since we need 3 to fill the 3 spots and we stop at 6 since after then we are simply going to overcount. Now we have 6P3. And we know that adding another element that is the same as one we have already counted will lead to overcounting, so we carefully add the last element which is by manually seeing how it would fit. 3 positions, 2 A are going to take 2 spots and the last element can be any of the other remaining elements, which are 5, 5 elements 1 spot, 5P1 = 5 And for these 2 A that we are manually finding the spots for, A _ A A A _ _ A A are the only 3 ways to put them in the spots, and since they are identical, A1 _ A2 and A2 _ A1 are the same thing so we leave them out and count 3 instead of 6. 3 ways to put the 2 A and 5 remaining elements so 5P1 * 3 which is 15, we already had the 6P3 which was all the permutations with just 1 of A and the rest 5 letters for 3 positions, and now we have the permutations for the next element which is an A and we have 6P3 + (5P1 * 3) = 135 This was for the 6P3 approach, but there's also the 5P3 approach, which is just saying we don't count the permutations with either A and find the permutations for the other 5 elements on 3 positions. Then we find 5P2 when 1 position is guaranteed A and lastly 5P1 when 2 positions are guaranteed A.
I came with a more difficult variant of this type of question. How many ways are there of a) selecting and b) arranging, four letters of the word "connection"? The answers which I arrived at were 98 ways of selecting along with 1960 for arranging.
I forget how I solved it back then but I wrote a Python code to solve the problem and got 149 for selecting and 758 for arranging. Here's the code: """"How many ways are there of a) selecting and b) arranging, four letters of the word ?""" from itertools import permutations, combinations def calcalate_comb(): """Calculates the number of unique four letter words of the word without regard to order""" letters = list("connection") #print(letters) #print(len(letters)) comb = combinations(letters, 4) total_comb = set() for i in comb: a, b, c, d = i word = (a, b, c, d) complete_word = "".join(word) total_comb.add(complete_word) print(total_comb) print(f"Answer:There're a total {len(total_comb)} ways of selecting a four letter word" " ") # b) def calculate_perm(): """Calculates the number of unique four letter words of the word with regard to order""" letters = list("connection") perm = permutations(letters, 4) total_perm = set() for i in perm: a, b, c, d = i word = (a, b, c, d) complete_word = "".join(word) total_perm.add(complete_word) #print(len(total_perm)) print(total_perm) print(f"Answer:There're a total {len(total_perm)} ways of arranging a four letter word") def main(): calcalate_comb() calculate_perm() if __name__ == '__main__': main()
@@ravenousturtle8498 Selection All different _ _ _ _ CONECTIO so 8C4= 70 With 2N N_ _ _N COECTIO so 7C2=21 With 3N N_N N COECTIO so 7C1=7 8C4+7C2+7C1=98 Arrangement All different _ _ _ _ CONECTIO so 8C4*4!= 1680 With 2N N_ _ _N COECTIO so 7C2*4!/2!=252 With 3N N_N N COECTIO so 7C1*4!/3!=28 8C4*4!+7C2*4!/2!+7C1*4!/3!=1960
This isn't 6P4. We are saying that there are 6 letters to choose for the first box, 5 letters in the second, and 4 in the last box, which is 6*5*4, which is 6P3 = 120 ways. We can kinda think about this way: There are 6 items to pick or choose, BUT we only want three of them, so we use the formula 6P3 for arrangements.
A bit late, but whatever lol. Since we eliminate the A's from appearing, there will be 5 letters to choose: {L, G, E, B, R} We only want three of them to appear and since order doesn't matter, we introduce combinatorics, nCr, which means that out of the 5 letters, we only want 3: 5C3
Yeah, personally, I don't really approach it that way. Consider this approach, which I think makes a lot more sense. For selecting, we don't care about the order. So, if we want to select three of them, here's how you can approach it. #1: Start by considering ONLY the individual or unique letters. In this case, there are: 6 letters {A, L, G, E, B, R}. Of these, we only need to select three of them: 6C3. #2: Now, consider the ways in which we can select a pair, meaning both A's. First, we need to understand how to select a pair. There is only one letter that we can find a pair from, A. So, there will be 1C1 way in selecting a pair. Now, from here, we only need to select one more letter, and it can be from any of the five remaining letters, so it'd be 1C1*5C1, which is 5. Adding them up, we get: 6C3 + 5 = 25 ways.
Here's how I learnt it from the video and processed it.
When there are lesser positions than the elements to pick or choose from we can’t simply divide by the overcounting. This is because it is not certain that for each arrangement there is the repeated element.
For ex.:
ALGEBRA and pick/choose 3.
But when we do 7C3/2! and 7P3/2!, we get a fractional number and that does not make sense as there can either be a possibility or none, not in between.
So to solve it we divide it into sub-arrangements and then add all them up.
For Combination,
First, no As: 5 total letters and 3 positions so 5C3 = 10
Then, 1 A: This A takes a guaranteed slot so only 2 positions left, 5C2 = 10
Then, 2 A: Similarly to 1 A, 5C1 = 5
And total is 25, which is our answer. As we see different types of combinations, ones without A, ones with a single A and ones with one more A as that’s the total no. of As, and adding them up are the total combinations for 3 positions.
Alternatively, we could directly just take 1 A and say ALGEBR, and in that case we just simply do 6C3, and this includes combinations with 1 A and no A. Which can be seen as it equals 20. Now we only need to see combinations with both As and when 2 A take up the slots, we are left with 1 position and 5 letters giving us 5 and total being 25.
Similarly, for Permutations,
First no As, 5P3 = 60
Then 1 A, 5P2 = 20
However, since order matters here, the 1 A’s place matters as well, as we know
A _ _
_ A _
_ _ A
are the 3 positions, we multiply 5P2 with 3
Then 2 A, 5P1 = 5
Similarly, these 2 A’s non identical positons matter as well and that gives us 15.
Total is 135.
Alternatively, we can directly take ALGEBR and that gives us 6P3 which is the sum of permutations achieved through 1 A and no As with order in mind, and it is indeed equal to 120 which we got from 5P3 + 5P2*3. And we follow the same last step which gives us 15 and the total is 135.
isnt there 7 letters? why pick 3 from 5
@@tokyo7671 That's because 2 of the A are repeating, as explained in the video and my comment, if we simply do 7P3 we are overcounting.
We know the 2 A in ALGEBRA are identical, and we need just 3 letters,
so if we do AAR or AAR, it is the same thing, and we have overcounted by 2.
Normally it wouldn't be a problem if we only had 3 letters and 3 positions, we can just divide by 2 leaving only 1 AAR as the only permutations would be AAR AAR ARA ARA RAA RAA and dividing by 2 removes half of them, which works because there are 2 duplicates for all elements. We aren't really dividing by 2, instead we are *not* counting the duplicates, which is a different thing because not counting means we don't include them in our permutations but division means removing them after they are generated. We can divide the final permutation count if we overcount by an exact figure, like 2 or 3 or so on and that works, that is to say division is a shortcut of reaching the same thing but it is merely a shortcut, not the actual method.
But we have 7 elements here, and now if we say 7 Pick 3, we have overcounted, but by an unknown amount, as if we simply divide it by 2 we will remove permutations that don't include 2 identical A. Like EBR or LRB etc., and simply halving them isn't guaranteed to take only 1 AAR of 2 AAR.
Assume the only permutations were LBR ERB AAR AAR GEB RBA then if you simply divide them by 2 you are saying you didn't count LBR AAR AAR, but as you can see, it always includes a permutation which shouldn't be removed, it is not an overcount. Then what do we divide the 6 permutations in the example to only take 1 AAR out of the 6 elements ? The answer is 1/6 which is a fraction, we can't really define a fraction, division was just a shortcut of saying we didn't generate x permutations and when it is a fraction, it is incorrect as you can either generate 1 permutation or none, but not 1.5 permutations.
So to solve this problem, we build up from a case where we can't overcount. We find all the permutations with 3 elements, then 4 ,then 5, and then 6. We start at 3 since we need 3 to fill the 3 spots and we stop at 6 since after then we are simply going to overcount. Now we have 6P3. And we know that adding another element that is the same as one we have already counted will lead to overcounting, so we carefully add the last element which is by manually seeing how it would fit.
3 positions, 2 A are going to take 2 spots and the last element can be any of the other remaining elements, which are 5, 5 elements 1 spot, 5P1 = 5
And for these 2 A that we are manually finding the spots for,
A _ A
A A _
_ A A
are the only 3 ways to put them in the spots, and since they are identical, A1 _ A2 and A2 _ A1 are the same thing so we leave them out and count 3 instead of 6. 3 ways to put the 2 A and 5 remaining elements so 5P1 * 3 which is 15, we already had the 6P3 which was all the permutations with just 1 of A and the rest 5 letters for 3 positions, and now we have the permutations for the next element which is an A and we have 6P3 + (5P1 * 3) = 135
This was for the 6P3 approach, but there's also the 5P3 approach, which is just saying we don't count the permutations with either A and find the permutations for the other 5 elements on 3 positions. Then we find 5P2 when 1 position is guaranteed A and lastly 5P1 when 2 positions are guaranteed A.
@@tokyo7671 China's better than Japan
W mans
My dumbass still cant get why the repitition of a makes such a big diff
You know combinations suck when even the teacher needs to read off the sheet :P
That... was... awesome.
@Alfy ALEX He did at the very beginning with the 6*5*4 (6P3 - but he says "six p four" by accident instead of "six p three")
I came with a more difficult variant of this type of question.
How many ways are there of a) selecting and b) arranging, four letters of the word "connection"?
The answers which I arrived at were 98 ways of selecting along with 1960 for arranging.
How did you do it?
I got 98 but only for n do you not have to treat o as a separate value too?
I forget how I solved it back then but I wrote a Python code to solve the problem and got
149 for selecting and 758 for arranging.
Here's the code:
""""How many ways are there of a) selecting and b) arranging, four letters of the word ?"""
from itertools import permutations, combinations
def calcalate_comb():
"""Calculates the number of unique four letter words of the word without regard to order"""
letters = list("connection")
#print(letters)
#print(len(letters))
comb = combinations(letters, 4)
total_comb = set()
for i in comb:
a, b, c, d = i
word = (a, b, c, d)
complete_word = "".join(word)
total_comb.add(complete_word)
print(total_comb)
print(f"Answer:There're a total {len(total_comb)} ways of selecting a four letter word" "
")
# b)
def calculate_perm():
"""Calculates the number of unique four letter words of the word with regard to order"""
letters = list("connection")
perm = permutations(letters, 4)
total_perm = set()
for i in perm:
a, b, c, d = i
word = (a, b, c, d)
complete_word = "".join(word)
total_perm.add(complete_word)
#print(len(total_perm))
print(total_perm)
print(f"Answer:There're a total {len(total_perm)} ways of arranging a four letter word")
def main():
calcalate_comb()
calculate_perm()
if __name__ == '__main__':
main()
@@ravenousturtle8498
Selection
All different
_ _ _ _
CONECTIO so 8C4= 70
With 2N
N_ _ _N
COECTIO so 7C2=21
With 3N
N_N N
COECTIO so 7C1=7
8C4+7C2+7C1=98
Arrangement
All different
_ _ _ _
CONECTIO so 8C4*4!= 1680
With 2N
N_ _ _N
COECTIO so 7C2*4!/2!=252
With 3N
N_N N
COECTIO so 7C1*4!/3!=28
8C4*4!+7C2*4!/2!+7C1*4!/3!=1960
it's still quite hard to understand
I will forever remember "you must think *thunk*" 4:58
Why do you say 6x5x4 is 6P4? Sorry I'm confused; isn't it 6P3?
A video on permutation of
numbers being divisible by 125 formed using 0,2,5,1,8 repetition allowed?
1000 is a multiple of 25 therefore any number where last 3 digits are a multiple of 125 is divisible by 125
Multiple of 25
125✓
250✓
375×
500✓
625×
750×
875×
1000✓
5 digit number
_ _ 125 (4*5 ways)
_ _ 250 (4*5 ways)
_ _ 500 (4*5 ways)
3*4*5=60 ways
_ 1000 (4ways)
4ways
60+4=64 ways
4 digit number
_125 (4ways)
_250 (4ways)
_500 (4ways)
3*4=12 ways
1000 (1way)
12+1=13 ways
3 digits
125 (1way)
250 (1way)
500 (1way)
3ways
Total permutation
64+13+3=80 ways
You're saving my life. I love you
My brain: confusion of the highest orda!!
@4:43
6P4 = 360
awesome explanation btw ! :)
This isn't 6P4. We are saying that there are 6 letters to choose for the first box, 5 letters in the second, and 4 in the last box, which is 6*5*4, which is 6P3 = 120 ways. We can kinda think about this way:
There are 6 items to pick or choose, BUT we only want three of them, so we use the formula 6P3 for arrangements.
Oiiiiiiiiiiiiiii
he meant to say 6P3
so hes technically still right
Oh man, your videos make me speechless
He is videos?! What? Eddie Woo is a human being! Have some respect!
Beautifully explained sir! Great job :)
the OG Eddie Woo
im getting murdered by a problem. 4 greeting cards, mary selects 3. what are the total selections? now the answer is 20 but i have no clue why
THANK YOU!!!!! this helped me figure out some problems !
i dont get it!!!! why did you choose 3 out of 5 samples in combination in your first example @2:50
A bit late, but whatever lol.
Since we eliminate the A's from appearing, there will be 5 letters to choose: {L, G, E, B, R}
We only want three of them to appear and since order doesn't matter, we introduce combinatorics, nCr, which means that out of the 5 letters, we only want 3: 5C3
still confused lol :"( gonna fail hsc gdbye
Yeah, personally, I don't really approach it that way.
Consider this approach, which I think makes a lot more sense.
For selecting, we don't care about the order. So, if we want to select three of them, here's how you can approach it.
#1: Start by considering ONLY the individual or unique letters. In this case, there are: 6 letters {A, L, G, E, B, R}.
Of these, we only need to select three of them: 6C3.
#2: Now, consider the ways in which we can select a pair, meaning both A's. First, we need to understand how to select a pair. There is only one letter that we can find a pair from, A. So, there will be 1C1 way in selecting a pair.
Now, from here, we only need to select one more letter, and it can be from any of the five remaining letters, so it'd be 1C1*5C1, which is 5.
Adding them up, we get: 6C3 + 5 = 25 ways.
For arranging, it's basically the same approach; the only difference is that we need to consider how many of these are arranged.
@@geraldhuang7858 Thanks, that was really helpful
can u post the question?
At 4:38 why are there 3 ways of doing them?
Because it's:
1) AAL or AAG or AAE or AAB or AAR
2) ALA or AGA or AEA or ABA or ARA
3) LAA or GAA or EAA or BAA or RAA
:)
China #1
Kickass delivery
I never understood this chapter
now where you are
@@tahabashir9405I do understand it better now! Thanks for asking
Its 6P3 by the way
why 3 dislikes?
why 13 dislikes
Why 22 dislikes?
Now we can't even see the dislikes :(