The Fundamental Counting Principle
HTML-код
- Опубликовано: 10 июн 2024
- This video shows how the Fundamental Counting Principle can be used to solve counting problems. This video was created for the Data Management (MDM4U) course in the province of Ontario, Canada.
It was 10 years ago but this is the best explanation I've ever watch. Well done!
Hey, thanks. This is one of my most commented on videos. Glad it was useful to you.
This is the best explanation of the fundamental counting principle I've ever seen. Thanks so much.
The video is very helpful for people who start learning the FCP. There seems to have a little glitch. It is in example 6.
252 seems wrong. It seems to be 256. The reason is as follows.
Suppose the last digit is a 2.
If the middle is 0, the first digit has 8 choices; if the middle is not a zero, it has 8 choices and the the first digit then has 7 choices which by the counting principle results in 56 cases. Altogether there are 64 cases when the last digit is a 2.
The reasoning is the same for 4,6, or 8 to be the last digit. This gives 64x4 or 256 numbers.
The final answer is 72 + 256 or 328.
The following R code confirms the new result.
count =0
for (a in 1:9){
for (b in 0:9){
for (c in c(0,2,4,6,8)){
if ((a-b)*(b-c)*(a-c)!=0) count=count+1
}
}
I understand your reasoning. I think the 7*9*4 I had, should actually be 8*8*4 = 256. Thanks for pointing that out. Have a great day.
Hi Shiju, I got the same thing as well with your logic on paper. The three cases are as follows:
(1) 0 in the third digit, 1-9 in second digit, 1-9 in the first digit other than second digit leaving 1*9*8=72 permutations
(2) 2,4,6, or 8 in the third digit, 0 in the second digit and 1-9 other than the third digit leaving 4*1*8=32 permutations
(3) 2,4,6, or 8 in the third digit, 1-9 in the second digit other than third digit, 1-9 in the first digit other than second and third digits leaving 4*8*7=224 permutations
The total permutations are 72+32+224=328 permutations. The trick was in the third case, where we had to exclude zero in the second digit, because we already accounted for a zero in the second case.
I just had to slightly modify the R code to include the multiplication operator between the parentheses in the last if condition. I am new to R, but here is the code:
count =0
for (a in 1:9){
for (b in 0:9){
for (c in c(0,2,4,6,8)){
if ((a-b)*(b-c)*(a-c)!=0) count=count+1
}
}
}
Whaou, i don’t know what to say. I m being struggling with this the entire semester until now. Thank you, thank you, thank you,..., thousands, millions of times... it s still not enough to thank you. But, you awesome... very very very well explain. I went back all over the exercises I was still not understanding, and this understanding makes things work better now. Thanks again.
Hi!! Wut grade r u??
Francine Salazar I’m a Junior what about u
@@xaviersmith2004 lol not u hahah I'm senior grade 10
Who hate math
Agreed
Pinoy grade8 4th quarter anyone??
same pre
advance kayo! tangina grade 10 pinagaaralan nmin ngaun lng...pero di yan ung main topic kundi permutation and combination...kaazar dahil basic lng daw yan hirap tuloy ako makapick up sa maintopic namin 😭
@@enzocruz9748 sadd, d2 sa JBLCF(school namin) advanced mag isip lahat
ayos lng yan mas better pag advance...yang fcp basic lng yan(sabi ng iba) permutation and combination need ng calculator sa sobrang haba ng numbers umaabot ng 1B hahaha...need nyo muna magets pano mag fcp para may idea na kayo tungkol sa permutation and combination
@@enzocruz9748 grade 10 here na may stat tas probability exam bukas. lol. permutation tas combination last namin
This video makes me think more, it has so many things to learn,Thanks
Pinoy grade 10 4th quarter anyone??
Yeah man
Yup lmao
grade 8 literally
Grade 8 fourth quarter 💀
*8
I like this video the teacher taught it how you did but you just explained it so much better
Great Tutorial Thank You!
thanks for continuing to upload videos! even though i might not use them now knowing that there is this awesome resource makes me a lot more confident when i come to reviewing a topic!
Woah u comment 8 years ago rlly old
Thank you so much!
i have learned in this video is how to solve a problem by using fundamental counting principle.
I've learned in this video is how to solve a problem by using fundamental counting principle.
Heck yeah me too
I've learned on this video is how to solved problems using the fundamental counting principle .
who payed you too say rhus
Excellent video - one minor point - for example 6b), the correct answer is 328, not 324. The math in the video fails to account for four combinations with a middle digit of 0: if the combination ends in -02, -04, -06 or -08, there are 8 choices for the first digit, not 7.
Oh so you mean 328 is the right answer?
@@elinacooper9627 Yes.
thanks man this video was super helpful :)
Thank you!!!
Thank you
In example 6, there are other options, for instance, number 2,4,6,8 as
the end number, we should divide the case into two pieces , one case is
that 0 is in middle, then there are 8 numbers we can choose as the first number, the other
case is there is no 0 , so there are 7 numbers we can choose as the first number. The
final answer will be 72+8*1*4+7*9*4=72+32+224=328 , what do you think?
For example 6b, if the middle digit is 0, then the first can have 8 possibilities.
I was confused by that too
Same question, do you think we should add where the condition is when the middle digit is zero?
@@johnpaultacsagon9941 actually we do that when the events are mutually exclusive events where the events can't occur at the same time
Example 6B can have 3 possibilites depending on whether two choices that we picked were even or not so if no even is selected for the first two spots then we can have 9*9*5, if we have even in the first spot then we can have 9*9*4, and lastly if both two spots were even then we have 9*9*3. Am i right or wrong?
Not sure where you are getting 9x9x5. If neither of the first 2 numbers are even, there are only 5 odd numbers (1,3,5,7 & 9), so it would be 5 x 4 x 5.
thank you
I'm sorry if my question sounds so stupid but in example 6 letter b, why does the last number ( the even ones) count as four and not one since only one number will be use each time? I'm sorry I'm just so dumb XD but I really don't get it
Hey, the only stupid questions are the ones not asked, so don't apologize for asking. The reason the last calculation has a ×4 at the end is that there are 4 different even numbers you could have as the last digit (i.e 2, 4, 6 or 8). The 0 was already taken care of as the end digit in the first calculation, so we don't have to consider the 0 being at the end for this one. Hope this helps.
I want a grilled cheese with a sundae
I'm with this guy, I want grilled cheese with sundae too
i have an exam tomorrow and i hope i pass it! thank you so much for your help :)
So how was the exam
We need answers lad
Thank you! This helped :)
Example 6
b) why there are only 7 digits left for hundreds place that is only possible when at tens place there can’t be a 0. But there can be a 0 at tens place. How would you justify this ?
There are 2 different cases shown. The top 8 - 9 - 0 is if the number ends in a zero, the one below is if the number ends in either 2, 4, 6 or 8. The 4 x 7 x 9 means there are 4 choices for the ones digit, 9 for the tens since the tens must be different from the ones (so 10 - 1 = 9) and the 7 is for the hundreds. Whatever is used for the tens and the ones can't also be used for the hundreds. Also, the hundreds has already had one digit used from the above example, so 10 - 2 - 1 = 7 for the hundreds. Hope this makes sense.
In the last practice problem, where the repetition of numbers is not allowed, why don't we do 7* 9 * 5 in one go where 5= all the even digits including zero, 9=all the other numbers except the one already used in the units, and 7 as the first digit (1-9 numbers and excluding repeats and can't be zero)?
Hey. The reason 7×9×5 doesn't work is the tents digit may also be an even number and it's not the same as the even ones digit. So, there aren't always 9 possible digits for the tens place. My solution of 324 is off by 4. Check our Joshua Parr's comment from about a year ago.
Our teacher doesn't even teach us and only give us activity to solve 😢 I only learn here
In example 6, how is 0 in the first digit not allowed? for example, if my security code on a credit card was 011, that would be a 3 digit number.
While you are correct, generally 011 would be considered a 2 digit number since it's the number 11. Security codes are an exception. You wouldn't call 000451 a 6 digit number since it's really just 451.
hey i honestly didn't get it why you counted separately for 0 and other 4 even number in number 6b. Is there any easy way you can make me understand since i went through all your comments yet didn't get it .
Hey, I'll try. Hope this helps make sense of why the 0 is handled separate from the 2, 4, 6 & 8.
The reason is that if a 3 digit number starts with 0 it's not considered a 3 digit number. For example 076 is really just 2 digits since it's just the number 76. If I talk about a case where the last digit is 0, 2, 4, 6 or 8, then it could end in 4 (for example) but the first digit could be 0, so it's only 2 digits then. When I say the last digit must be 0, then the first digit can't also be 0.
Thank youuu
In 13:09, is it impossible to have odd numbers as last digit? Why, I'm so confused because of that part.
+Paula Denise Palencia Hi. It's because the number is supposed to be even, which means it can only end in 0, 2, 4, 6 or 8.
Oh, thanks, I didn't notice that it's about even numbers. HAHAHHAHAHA. Thanks, anyway :)
There are total of 328 3-digit non-repeating even numbers.
Please tell me how😩 I tried his way but it kept giving me the wrong answer compared to the one in the textbook!!! What way do you do it?
please answer that
a)if repetition are not allowed how many three digit numbers can be form the six digits 2,3,5,6,7,9
b) if less than 400
c) multiples of 5
how to solve that thanks
i need your help ?
a) if there are no restrictions for a) then since there are 6 numbers, we multiply 6X5X4 = 120
b) if the number must be less than 400 it can only start with 2 or 3, so the first digit can only be filled in 2 ways, the second or third digit can be any other number not already used. So we multiply 2X5X4 = 40. The second number if 5 since one number of the 6 has already been used and the 4 on the end is because 2 of the 6 digits has been used.
c) for multiples of 5 the last digit must be 5 because no number ending in 2, 3, 6, 7 or 9 is a multiple of 5. So the first number can be selected in 5 ways (from 2,3,6,7 or 9). So we multiply 5X4X1 = 20. The second number is 4 since the last digit must be 5 and a number has also been used for the hundreds digit at the beginning, so 6 numbers - 2 numbers used = 4 numbers.
Hello! On the last example, why is it 7 on 4*7*9? What if the digit 0 is placed on the middle box? Is it supposed to be 8?
Actually the answer for the last one should be 328, not 324. You'll see this in some of the comments below.
: The top 8×9 = 72 counts how many numbers end in 0.
: Next, if the last number is one of 2, 4, 6 or 8 then consider when the tens digit is 0 or isn't (2 cases). If the tens digit is 0 there are 4 ways to select the ones digit and 1 way to select the tens digit leaving 8 ways to select the hundreds digit, so 8×1×4 = 32.
: Lastly, if the ones digit is one from 2, 4, 6 or 6 and the tens digit can't be 0, there are 8 possibilities for the tens digit and there are 7 digits possible for the hundreds so 7×8×4 = 224.
Then 72 + 32 + 224 = 328 possible numbers.
Hope that makes sense.
@@AlRichards314 thankyou for the responseee! I will have a test tomorrow and your video helped me a lot :))))
@@AlRichards314 I also joined the conversation to point out that I got 328 for the number of outcomes. I have great respect for how you accepted the question/correction.
Your video has such high quality in presentation style and content. Very impressive! You are helping a lot of people. Have a great weekend.
@@barrydavis331 Hey. Thanks for the comment. You have a good weekend too!
A company makes trucks in 6 different sizes. Each size comes with a choice of 5 exterior colors, a choice of 3 interior colors, with or without CB radio, and with or without an extra horn. How many different types of trucks does the company make?
Hey. Each truck has every characteristic you mentioned. Thus we multiply 5×3×2×2 = 60 possible different configuration for this truck. The two 2s for the CB radio or extra horn represent that there are 2 possibilities for each: have the feature or don't have it. Hope that makes sense.
thank youuu❤️
I had misconceptions on the License plate number example.
in 6b, why do you need to separate the 0 from 2,4,6,8?
0 is handled separately from 2,4,6 & 8 because the first digit isn't permitted to be 0. So, if we consider the case where the last digit is 0, then (in that case) the first digit automatically isn't 0.
Can you check example 6 part b. again when the middle is ZERO and another for the last can be EVEN? I think the hundredth digit is different. Thank you
Hi. Sorry to take a whole day to reply, but I was away yesterday. I had enough people asking about 6b that one day I actually wrote all these numbers out to verify that it was 324 possible numbers.
I think if you are looking at when the middle digit is 0, that's just a different way of looking at the solution. I think that's considerably longer since you'd also have to consider cases where the middle digit is 1 or 6 or 9, etc..
@@AlRichards314 In the case that the last digit is even, If the middle digit is 0 the first must be any of 8 numbers (1-9 except the last digit and 0 that cannot be for the first) so we have 8 x 1 x 4 = 32 possible numbers. If the middle digit is non zero (1-9 and except the last digit) the first can be only any of 7 numbers (not the same as the last and the middle). So we have 7 x 8 x 4 = 224 possible numbers for this case. Therefore we have all three cases = 72 + 32+ 224 = 328 possible numbers. Please correct my method if I miss something. Thank you.
@@wutthichaimisong6637 Let me check with a colleague and I'll get back to you.
@@wutthichaimisong6637 I believe your answer is correct. It should be 328, not 324.
I approached it another way and also got 328.
I said the last 2 digits can either be even, even or odd, even. For the even, even there are 20 ways for that to occur since 4 x 5 = 20. Of these, when one of the 2 even digits is 0, there will be 8 ways to select the hundreds digit and that happens 8 times of the 20. The other 20 - 8 = 12 times there are always 7 possible digits for the hundreds digit.
If the tens & ones digits are odd, even, there are 5 x 5 or 25 ways for that to occur. When the last digit is 0 there are always 8 ways to select the hundreds digit and that happens 5 times (1,3,5,7,or 9). So the other 25 - 5 = 20 ways there are always 7 ways to pick the hundreds digit (1,3,5,7 plus 3 other even numbers).
So, altogether, there are 13 way with 8 possible numbers, plus 32 ways with 7 possible numbers. 8 x 13 + 7 x 32 = 328.
Yes, counting problems can be somewhat complex. Hope this makes sense.
@@AlRichards314 Thank you! :)
Watching after 9 years
A set of cards consisting of three D's two O's and one x.find the number of ways of getting the word DODO. Can u solve this problem sir
Hey, interesting problem. This involves identical objects. Notice all 3 D's are exactly the same and the two O's as well. Someone might think there are 3 ways to select the first D, times 2 ways to select the first O, etc. But since all 3 D's are the same there's only 1 way to start with D and both O's are identical, so there's only 1 way to get DODO.
at the last example with no repetition, why is the first digit's number of possibilities only 7? shouldn't it be 8?
The top scenario includes the number 0 (so 0 to 9), but the bottom scenario doesn't include 0 (so 1 to 9). Since this only uses 9 numbers and the number on the end is already taken and the tens digit in the middle is already used, then 9 - 2 = 7 numbers left we could put in the hundreds digit.
@@AlRichards314 ohhh, thank you so much!!
Brother i m not clear still. Actually two numbers mean two digit number . That is 5 and 6 like that? .why do u use 10 power no of multiples?
I'm assuming you are referring to example 4. The 10² is from the 2 digit part of the plate. There are 2 numbers and each can be selected in 10 different ways (0,1,2,3,4,5,6,7,8,9). Thus 10 × 10 = 10².
In example 6, there are other options, for instance, number 2,4,6,8 as the end number, we should divide the case into two pieces , one case is that 0 is in middle, then there are 8 numbers we can choose, the other case is there is no 0 , so there are 7 numbers we can choose. The final answer will be 72+8*1*4+7*9*4=72+32+224=328
Thanks for the comment. I agree with your 328 answer. If you look down through the comments to 5 months ago, you'll find the 328 discussed with another viewer.
Roses are Red
Violets are Blue
If you learned a lot,
Make that like button blue
Notebooks come in 4 colors: red, yellow, green, and blue. They come in two types: 5-subject and 7-subject.
Pls explain what happen to example 5?
The tree diagram is a way to organise to properly count all possibilities. The number of possible outcomes are listed on the right side of the diagram. There are 3 ways to pick the president times 3 ways to pick the Treasurer times 2 ways to pick the secretary. So there are 3X3X2 = 18 possible outcomes.
wow
thank you
?hello , thank you but i dont understand Example 5 , can you explain it
In example 5 there are 3 ways to pick the president (since it has to be a female this year). There would be 3 ways to select the treasurer (Bruce & or one of the 2 remaining females) and there would be 2 people left that could be secretaries. So, altogether there are 3x3x2 = 18 different groups that could be chosen.
How do you get the 10?please answer
Do you mean the 10 in example 1? The main course can be chosen in 2 ways and for each of those there are 5 possible desserts, so 2×5 = 10.
Thank u dear
If arrangment of the plates only two letter like
AAB
DDE
CCT
ASD
RAY
How many number
Do you mean exactly 2 alike or at most 2 alike? AAB, DDE and CCT all have 2 letters the same, but ASD and RAY have none the same. If 2 letters were the same, there would be 26 ways to pick the pair that are alike and 25 ways to pick the one that is different, So, for exactly 2 alike there would be 26 x 25 = 650 of these 3 letter arrangements (which is also P(26,2).
If all the letters are different there would be 26 ways for the first, 25 ways to pick the second and 24 ways to pick the third letter. So, for all three letters different there would be 26 x 25 x 24 = 15,600. The is also P(26,3), which also equals 15,600. Hope this helps.
@@AlRichards314
شكرا جزيلا
Thank u very very mutch for quik replay
My qs is i have 4 letters
A B C D i want make all words by 3 letter
The letter repeat twice such
AAB
DDC
DCD
BBC
and also include other words such
ABC
DAB
I try manually i found 60
I e 4 power 3 minus 4
If there is formula
I hope u understand what i need
Thanks
@@abousamah1967 The number of 3 letter arrangements from 4 letters where all letters are different would be 4x3x2 = 24. If 2 letters are the same, the 2 letter pair could be chosen in 4 ways x 3 for the other different letter. So if 2 letters are the same that can happen in 4 x 3 = 12 ways. If all 3 letters are the same, there's only 4 ways for that to happen. So there would be 24 + 12 + 4 = 40 possible arrangement.
Now, if you consider AAD different from ADA, that will increase the number of possible arrangements.
@@AlRichards314
thanks for your interest
my dear
this example such cars plates in my country
AAB/ABA/ABC/TTA/TTP/TYH
not allowed AAA ONLY AA
finally i get this formula
3 letters word from 4 letters ( 4 power 3)-4=60
3 letters word from 5 letters ( 5 power 3)-5=120
3 letters word from 26 letters ( 26 power 3)-26=17550
please check whether right or wrong
thanks
@@abousamah1967 So, you just can't have all the letters the same? If that's the case, then your last 3 formulas are all good.
i still dont get the repetition part
With repetition a digit or character can be used multiple times (or even in every spot). For example if you were arranging 3 letters, you could even have BBB if the B was used every time. So with repetition there are 26 choices for each spot so there would be 26x26x26 or 26³ possible "words". If repetition wasn't allowed, then every character must be different, so there would be 26x25x24 possible "words".
The repetition example had two other complications since it was supposed to be an even number so it must end in 0, 2, 4, 6 or 8, hence why there were 5 ways the last digit could be filled. It was also to be a 3 digit number, so the hundreds digit couldn't be 0, hence why the first digit must be a number from 1 - 9. Hope this helps clarify.
for example 5, why wouldn't it be 4x4x3 ?
I assume your 3 is because there are 3 females, so that's fine. It's not stated in the problem, but one person can't assume 2 different roles. So someone can't be the treasurer and also the vice-president. So, if one of the females is selected as the president, then there are 2 females and one male left for the vice-president position (so that's 3 people). Once those 2 roles are filled, there are 2 people left for the remaining position, hence why it's 3×3×2 = 18.
Anyone after 10 years later in 2022.
on the repetition of digit is allowed why did you do it twice???
pls reply????
Do you mean ex 6 with a & b? or in a) when it shows 9X10X5?
yeah on the ex 6 b
mistaken
on the repetition of the digit is not allowed
Sorry, it's taken so long to respond, but we have some company the last few days. I did this in 2 cases. The first one was if the even number ends in 0, and the second way is if it ends in 2, 4, 6 or 8. The reason you do this separately, is that a three digit number can't start with 0, like 058, because that's really only a two digit number.
thank you for the answer
I have one question again
a five cards are drawn from a standard deck of cards three are block and two are red??
a:how many if the repetition are allowed and if it's not allowed????
answer pls???
thank you.....
I TILL DONT I UNDERSTAND EXAMPLE NO. 6 THE REPETITION OF DIGIT NOT ALLOWED
HOW YOU SAY THE 0 TO THE LAST DIGIT WHY?
Since the three digit number is supposed to be even, it has to end in either 0, 2, 4 ,6 or 8. If it ends in 0, then the ones or tens digits can be 2, 4, 6 or 8 or if the last digit is 4, then the tens could be 2, 6 or 8 for example.
I'm not 100% sure if I am answering your question. If not then let me know more clearly what you are asking.
umm..
if the the question involve the zero number like your example and even number repetition not allowed
that was use a case1 and case 2
that was my question in this senaryo?
The 0 can't be in the first digit since that makes them not a 3 digit number. For example, if you write the number 043, that's not a 3 digit number because it's really just 43. So, in both examples, the first digit (hundreds) has to be only numbers 1 through 9 (and can't be 0).
how many 3 digit even number can be created if repition not allowed ?
plz answer that thanks?
cherry-ann Herras 324 possible numbers
🙂
Hi Teacher 😎😎
I don't understand example six, aren't there only 4 even numbers?
The even numbers are 0, 2, 4, 6 and 8, so there are 5 of them.
Thank You, I missed the fact that an even number ends with an even number..
You're welcome. Have a nice day.
@@AlRichards314 please help
The question says 3 digits even number so I thought all three digits were going to be inform of even number but to my surprise its only the last digit while the first two are subjected to all digits
@@gracebwakya5493 A number is called "even" if it ends in 0, 2, 4, 6 or 8. The other digits can be any number from 0 to 9. For example 26 is even, but 58 is also even, even though the tens digit is 5 (which is odd).
i dont understand the last bit the repetition and no repetiton
15:28 our lecturer said the answer would be 328-without repetition, I'm confused 😖⁉
Hey, yes 328 is correct. See the comments from 5 days ago or 6 months back. Sorry for the confusion.
@@AlRichards314oh okay thank you for your kind comment.
Thanks for the video. However, I am still struggling to understand Example 6 part B. Please explain more in detail. Looking forward to it .thanks
I'll try. Let me know if this helps. The last digit must be either 0, 2, 4, 6 or 8 in order for the number to be even. Also remember that the first digit can't be 0 since that would make it a 2 digit number (and not 3 digits), such as 076 is actually just 76 so it's not a 3 digit number.
So, the 2 cases are if the last digit is 0 and the other is if it's 2, 4, 6 or 8.
For the first one if it ends in 0 then the other digits must be selected from 1, 2, 3, 4, 5, 6,7 , 8 or 9. There are 9 choices for the tens digit and then only 8 left for the hundreds digit, so 9 x 8 x 1 = 72.
For the other case the number ends in 2, 4, 6 or 8 so there are 4 ways to pick that part. You've used 1 digit already so there are 9 left for the tens digit and then only 7 left for the hundreds since you've used 2 already and the hundreds can't be 0. So 4 x 7 x 9 = 252.
Then adding 72 + 252 = 324.
@@AlRichards314 Thanks for the quick response. Then why cannot I just use this equation: 1st digit 9 possibilities, 2nd digit also 9 possibilities, 3rd digit has 4 possibilities: 9*9*4=324. Is this approach wrong?
@@vusalmammadov1161 I see that 9×9×4 = 324, which is the correct answer, but the last digit has 5 different possibilities: 0, 2, 4, 6 or 8.
I Still don't understand Example 6. Can you please explain?. How many 3 "EVEN" number can be created"? a) repetition is allowed.
SO I'm thinking;
Even number between 0-9 are (0,2,4,6,8)
so, answer should be 5*5*5 = 125 (because there are five even numbers an can be selected that numbers of time)
I"confuse can you please explain where the 0 -9 is coming from when 9 is not an even number?
Thanks.
The number on the end must be 0,2,4,6 or 8 so there are 5 ways to select the last digit. In order for the 3 digit number to be even only the last or ones digit has the 0,2,4,6,8 restriction. The tens digit can be anything from 0 to 9, so there are 10 ways to select the tens digit. The hundreds digit can't be 0, but can be anything else from 1 to 9 so there are 9 ways to select the hundreds digit. Hope this helps.
B.com part i statistics k 8 .13
I’m still confused on the multiplication my question was 3 meat 3 veggie and 2 dessert the answer was 18 and did the multiplication and got 24: here’s the thing I did the long way but can’t do that for the huge ones: so either my multiplication was wrong or my school is wrong:I WAS WRONG I put 4 veggies when it was 3
What exactly was the question? 3×3×2 = 18, but I'd like to know what the question was exactly.
The answer is 328
Case 1: First digit is zero, then 8 x 9 x 1 = 72
Case 2: 2nd digit is zero, then 8 x 1 x 4 = 32
Case 3: First digit 2, 4, 6, 8, then 7 x 8 x 4 = 224
Total = 328
Hey. Thanks for the comment. I realized that some time after the video was upload. For example, see Joshua Parr's comment a year ago.
i don’t get it
thats sad
Same
LSGH Grade 9 4th Q?
goat
Thanks. Not often I'm in the same category as LeBron.
Mahal ko si J
How about 11 dresses, 6 bags and 7 shoes? How many outfits can a girl make?
Hey, as long as the girl's wearing 1 dress, carrying 1 bag and wearing 1 pair of shoes, she would have 11 x 6 x 7 = 462 different possible outfits with no regard for matching any colors of course. :)
I don't understand the example 6
Hey, there have been several good comments below about example 6. Take a look, they may help.
Hi, the answer is not correct for example 6 ,
engineer gaming
You can put ads to a dawn 10 year old vid. This vid is older than my brother aint no way it has ads.
no. 6b Is wrong
Why do you think 6b is incorrect?
AlRichards314 I think it must be 8 8 4
@@blank_ec828 Hi. 884 is way too high. I took the time to list them all, just as a check. When you are writing these numbers you can never have two or all three digits the same. For example, you can't have anything like 220 - 228, 330, 332, 334, 336, 338, you also can't have anything higher than 986, since all those numbers have 2 digits the same (990, 992, 994, 996, 998). My list has 324 numbers.
I think its literally my 10th time that Im watching videos about counting and still I can't get it
if there's a specific question, leave a comment and I'll try to help.
@@AlRichards314 thank u so much you are such a kind teacher 🌺🌺🌺🌹🌹🌹🌹I got it from your video ❤️
@@dragonartgroup6982 Hey, thanks for the kind words. Warmest regards.
Ol classes anyone? Yawa
Yawa mo lang hahahaha
Yawa😆 i have online school rn😂😂
Example 6. B) answer: 328
I'm Asian :)
Josh Joshua lucky
i hate math
Hey, very sorry to hear that. Was there something that needed a little more of an explanation?
Ok then
Thank you so much!