GMAT Permutation Combination Probability Basics | Part II | Free GMAT Online Classes
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- Опубликовано: 30 сен 2024
- GMAT Preparation Online | Free GMAT Online Classes | GMAT Permutation Combination & Probability Part 2
It is a sequel to the 1st session. Watch Part 1 @ • GMAT Permutation Proba... before watching part 2.
What is covered in this session?
1. We start by recapping key points from session 1.
2. Classify repeated sampling in a 2 by 2 matrix categorizing on the basis of whether order matters or not and whether the sampling is done with replacement or otherwise
3. What is permutation? Derive npr with an example
4. What is combination? Derive ncr with an example
5. Establish a standard framework to solve questions that fall under the sampling without replacement category.
6. Sampling with replacement and where the order matters.
7. Few simple examples to consolidate the concepts.
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This kind of teaching requires passion. I'm so glad I found this channel. Thank you sir!
Many thanks for your wonderful feedback!
Best wishes for your GMAT!
@@Wizako Thank you.
I wish i had found your course earlier.I took e-gmat coaching, but your teaching is better. Sir please do better marketing for your coaching bcoz its better than others! Thank you sir!
Honestly, thank you so so so much for this series. Can't articulate my appreciation enough!
Hello Eid
Thanks so much for your feedback. Did you watch the Quant Getting Started series of 2 videos. In case you have not watched those videos, you can watch them @ ruclips.net/video/1oHjbTz6_zs/видео.html and ruclips.net/video/4R4eFoti1Zc/видео.html
The answer to the Q1 regarding the sum of numbers rearranged (2468)= 133320
That's absolutely correct! 🙌
After getting a 55% in 11th standard Math and avoiding it for more than 13 years or my Life: after more than 6 years of work experience, working in 5 different countries, 2 degrees and 4 promotions. I have finally UNDERSTOOD Permutation!!!!!! FINALLY!!!
Combination is still to come; I'm watching the video.
Sincere thank you, Mr. K S Baskar Sir.
Hello Somendra,
Many thanks for this comment. Really made our day :)
Wishing you the very best! 🚀
My favorite math teacher ever. Nobody explains math in such an interesting and understandable way. Thank you very much.
Thanks for your feedback.
To appreciate your effort, I saw the whole ad and clicked on it too hehe
Hi Baskar Sir,
It's like even if a person doesn't have any doubt, he/she is bound to appreciate/comment in this or any video lecture of yours because they are way too good.
You ended the lecture with an important message that I literally jotted them down in the first page of my note-book:
1. Start somewhere. Do not fall prey to analysis paralysis.
2. Relentless practice.
3. Commit to the test.
Thank you so much.
Shubham.
Hi Shubham,
Thanks for the feedback!
I am glad you jotted the information down because often times, we lose track of what inspires us as well, when there is a snowball of information out there!
Make sure you reserve a fixed amount of time each day and that you structure your preparation. That's the surefire way to crack the GMAT!
Best wishes
My solutions for the 2 problems regarding sum of 4 digits.
Ans. 1 - 133320 [for the 2468 problem]
Ans. 2 - 53328 [for the 2248 problem]
Are they the correct answers?
For 2468, it is 66660.
@@mishraaparna how did you get it, cuz i am also getting 133320
@@kanakheda7156 I think my mistake, she multiplied all the places by three instead of six. In the beginning, even I did the same mistake. Only later I got the correct answer as 133320 (after multiplying 2+4+6+8 with six)
I got 62216 for the 2248 one, how did you get 53328 ?
CAN YOU PLEASE EXPLAIN THE CASE FOR 2248 PROBLEM?
133320 and 53328. got the two answers. however i would like to clarify more on the concept of the 53328 ques. i have almost 90% clarity on that question (thanks to ur brilliant teaching method), but if you share some more logic on then later ques, i will be obliged.
I got the same answer
Sir, this was one of the best perm and comb class I have ever attended. Thanks
Thanks Rose for your feedback.
I am not sure whether you have watched the remaining videos in the getting started series. If you have not, please watch these videos to get started with your GMAT quant prep @ ruclips.net/p/PL3JsYBL14ZltnwdqUt7NTyx9wUzVh-OZu
@@Wizako how is the logic for cards example? If I put card back, I don't have it. So at end doing 3 times I'll have only one card but I need 3 as you said. So this can be said to password example also. If I pick a letter and put it back I'll have only one. So it should be without replacement. Pls explain that.
the last few questions do not match the 4 steps that you told before
Best video for GMAT prep that I've watched till now, thank you!
Glad it was helpful Manasi!
Have you watched - "GMAT Quant Basics - Part I - Getting Started" ? If not, don't miss it! Watch it here: ruclips.net/video/1oHjbTz6_zs/видео.html
Best wishes for your GMAT prep!
nobody better at math and computer science than my indian brothers
Hello Zim
Thanks for your feedback.
Great insight to the topic.This was rushed by my GMAT tutor.
Thank you Bhaskar for sharing this, you are the best!
Glad you liked it!
@@Wizako I have completed the four videos from introduction playlist and I’m impressed with your way of teaching, interested to enquire about full length teaching lessons
@@riyavishwakarma3701 Can you call us or ping us on +91 95000 48484. Will be happy to walk you through our courses.
Answer for the first question-133320
Answer for the second question-53328 ? Can someone please tell me if this is correct?
Can you clarify which questions you are referring to? All answers to the questions are in the video and the first question discussed (numbers & digits 1) is about arranging three-digit numbers and the answer is not a 6 digit number! Please check or clarify
Thank you Wizako. You are simply heaven-sent. May The Lord of Heaven's armies bless you.
Wow, thank you! Thanks a ton for your feedback!
Happy learning and best wishes!
apart from being a excellent teacher mentor I want to compliment your sense of humour is Sir. you make sound maths so fun and easy. all maths teachers I got in school made the concept mostly more difficult.
Haha..! 😆
Many thanks for your interesting comment. Really appreciate it.
Cheers!
I did not get answer i am just started knowing anout GMAT started with sir lecture do i need to put all possibilities so that i can calculate repetitive numbers in units and 10 and 100 and 1000 place or any other way around can any one tell me ?
Hi,
Four digit numbers that can be formed = 4⋅3⋅2⋅1 = 24 ways we can form a four digit number.
Since it's a 4 digit number, each digit will appear 6 times (i.e, 24/4) times in each of units, tens, hundreds, and thousands place. Therefore, the sum of digits in the units place is 6(2+4+6+8) = 120
Similarly sum of digits at tens,hundreds,thousands place is 120.
Required sum of all numbers = 120 + 120⋅10 + 120⋅100 + 120⋅1000
Cheers!
Thankyou sir, You are such an elaborate teacher. Enjoyed every bit of it.
Glad to hear that Victoria.
Happy learning!
Thank you so much for these videos, they are the best ones out there! I was feeling really overwhelmed by the GMAT exam, and your videos have changed everything for me, so thank you!
133320 answer fr 2468
Spot on Rishi
Absolutely right. The key to solving this question is figuring out the number of times each of the digits repeat in any one place. You got it right.
Thanks to the Professor for this series. Counting and probability is my weak point and I am more comfortable with them now.
Glad you found our content helpful!
Cheers & Best Wishes for your GMAT Preparation!
Sir, I'm always getting confused in finding no.of ways
Hello Dhanush,
Would recommend revisiting the First Part to consolidate the concepts before you proceed with questions of higher difficulty.
You can watch it here - ruclips.net/video/A0MbCayeAAY/видео.html
Keep at it, practice, and you'll get there.
try @ home 1.) 133320; 2.) 53328 for 1:04:20 time numerical. Kindly suggest if it's correct.
Both answers are correct. Well done!
How is 53328 the ans???
Wizako you're the best! Thank you so much!
Hi Oni,
Very happy to hear this 🤗🤗
Many thanks for the comment 😍
Best wishes for your GMAT Prep!! 🚀
In how many ways can the letters of the word GRAPHITE be rearranged such that the positions in which the vowels appear do not change? - The answer should be simply 5!. But if the question was - "In how many ways can the letters of the word GRAPHITE be rearranged such that the Relative Order in which the vowels appear do not change?", the answer would be 3!x5!, that you showed in the video. Please correct me if I'm wrong.
Hi sir in scenario 3, how can the same student be both president and vice president?
The question states a person can hold more than one post. So, it is a case of sampling with replacement.
Wow .. Really Organised and I guess I dont need to refer any book or anything less. Just confident to practice . Appreciate Sir such Free sessions really made a great differnce for society . Keep doing excellant to world
Thanks Ankit for your feedback. Did you watch part 1 of this series? In case you have not yet watched it, please watch it at ruclips.net/video/A0MbCayeAAY/видео.html. The first part helps get definition in place and provides clarity to the idea of replacement and ordering.
Yeah I have watched almost all your videos available RUclips or Q51 series .
I have total 5 years of work experience and working with Cisco organisation . I want to grow further in my career and I am really impressed with your tips. Thanks a lot sir
Are you planning to do a 2 year MBA or a 1 year MBA? If it is 2 years, the sweet spot is applying when you have 4 to 7 years experience. If it is one year, and you are looking at the IIMs for their PGPX, apply when you have 7 to 10 years experience.
Question for How many 3 digit numbers that do not contain 3 as one of their digits. I calculated in a different way and I am not getting the same answer. Why can;t we take this way?
1st sample- hundreds digit has 3 and the rest 2 doesn't => 9 X 9 X 9
2nd sample- tenth digit has 3 and rest 2 doesn't => 8 X10 X9
3rd sample- ones digit has 3 and rest 2 doesn't => 8 X9X 10
Add them all and then we know these have 3 in atleast 1 one if them and not in other 2. This sum is way greater than our answer. What am I missing?
59:47
133320
I am amazed by your question types, thank you so much for making this video.
Hi Cornelius,
Thanks for the feedback.
You can also check out part 1 of the video here - ruclips.net/video/A0MbCayeAAY/видео.html
How is 53328 the ans for 2248 que???
The number of rearrangements for 2248 is 4!/2! = 12
So, 3 of the 12 numbers will end in 4, another 3 will end in 8, and 6 of the 12 numbers will end in 2.
The sum of the units place is 3*4 + 3*8 + 6*2 = 48
The sum of the tens place is multiplying the above sum by 10 = 480
The sum of the hundreds place = 4800
The sum of the thousands place = 48000
The final sum is the sum of the units + tens + hundreds + thousands = 53,328
For the first time ever I could understand the concept of P&C which I have always been afraid of until I came across your video lessons. You are really amazing sir. Hats off to your dedication.
Thank you very much for your feedback. Not sure whether you tried these questions from the playlist @ ruclips.net/p/PL3JsYBL14ZltoiZSpm5LCDjlYLj7r7k24
Hi Sir! Referring to minutes 51:30, how do I find that 10C3 is 120 without a calculator in the exam? Even with the formula = n!/[(n-r)!r!], I can't understand how manual calculation is used. Please help!
Thank you in advanced for your good work!
Hello Paul
Please use the following steps to compute the value of 10C3 without using formula
1. Product of 3 numbers counting downward from 10. i.e., 10 times 9 times 8 = 720
2. Product of the first 3 numbers i.e., 1 * 2 * 3 = 6
3. 10C3 is step 1 / step 2 = 720/6 = 120.
Let us take another example. Value of 17C4
Step 1: Product of '4' numbers counting downward from 17.
17 * 16 * 15 * 14 (you will be better off not computing this value as it will cancel with terms in step 2)
Step 2: Product of first 4 numbers: 1 * 2 * 3 * 4
17C4 = Step 1 / Step 2 = (17 * 16 * 15 * 14)/(1 * 2 * 3 * 4) = 17 * 2 * 5 * 14 {cancelling 16 with 2*4 and 15 with 3}
= 170 * 14 = 2380
Please pick 5 to 10 numbers and run through this process. You will be able to compute the value in less than 30 seconds after that practice.
Cheers
@@Wizako Thank you for the help! 😄
how is the logic for cards example? If I put card back, I don't have it. So at end doing 3 times I'll have only one card but I need 3 as you said. So this can be said to password example also. If I pick a letter and put it back I'll have only one. So it should be without replacement. Pls explain that
Hi Sir,
Kindly guide me in solving the below question
Question : Ten people are to be seated at a rectangle table for dinner, Tanya will sit at the head of the table. Henry must not seat beside either Wilson or Nancy. In how many ways can the people be seated for dinner.
Thanks in advance.
I got it ! :)
@@moumitadey2363 Hello Moumita!
I found this Q interesting, I am curious, what was the answer for this Q?
My God! It is such a fantastic video!
Hi,
Many thanks for the feedback. Really appreciate it. :)
Wishing you the best for your GMAT prep!
Just amazing... much thanks
Glad you liked it! Good luck with your GMAT. Feel free to reach out to +91 95000 48484 to attend a free class or check out more content as a trial user.
Thanks a lot sir for your videos. You have made P&C very easy to understand. ☺️☺️
Hello Shobhit
Thanks for your feedback. Few more questions on permutation and probability that you can try @ ruclips.net/p/PL3JsYBL14ZltoiZSpm5LCDjlYLj7r7k24
Thanks a lot for making this incredible video lecture.
I have one questions regarding the sport committee and music club example. I do not quite grasp the concept that how does the order matter in the case if one member occupies both posts. How is the order of any significance when one guy is occupying the same post i.e while listing down the possibilities we will write for both scenarios as XX (X being the person selected to do the job).
Can you please explain this concept a bit more.
Thank you sir. You are very much appreciated ♥️
Glad it was helpful!
Thank you very much for the lesson, I have spent a couple of hours trying to resolve Dorm Warden problem. I still don't understand what is n and what is r for the Step 2. I figured that the number of ways of reordering would be 4!*4!*4! Am I going in the right direction? This topic of permutations is completely new to me, would it be possible to guide me through this problem? Many thanks in advance
i think i must ve seen atleast 20 videos... wrt npr and ncrs ....i think its a blessing that i am able to learn from u...so clear and simple made my basics much better .. thx bhaskar sir🙏🏻
Thanks a lot sir.. very easily explained...one word ...best
Thank you so much sir. Your videos are brilliant 🙏
Hello Katherine
Thanks for your feedback.
We have 4 videos in the getting started series. You could access them by watching this playlist - ruclips.net/p/PL3JsYBL14ZltnwdqUt7NTyx9wUzVh-OZu
Nice trick, thank you
Hello Vishwajeet!
Thanks.
Best wishes for your GMAT prep!
Stuffs are really helpful
👍
Thank you very much sir. Appreciate it. Permutation and Combination always scares me but the way you have explained in your video makes me want to learn the subject more.
m taking GMAT soon . this is one class which taught me something about P&C . Till now i was a complete dumb about the P N C .
Hi Pawan,
No worries! :)
All the best for your GMAT!
Your explanations are always very helpful! Your advice at the end is also really appreciated. Thank you!
kind sir, Please post probability basics also. please. your content is best .
You have been my 1st person to go to for any math issue for gmat appreciate your hardwork...will be having a bright future only because of you..thank you☺☺☺❤🙏
Hi Kirti,
Very very happy to hear this!! Best of luck for your GMAT preparation!
Heart of this video is at 39:00. This snapshot just shows the quality of teaching, expertize and command on subject you have Sir. Thank you for putting down this video to general public. I have been struggling a lot with this topic and after watching this video it just cleared many of doubts and boosted my confidence.
Glad it was helpful!
Hello sir, can you please explain the 4 digit sum problem, I am stuck in number of ways of Reaarangements
Hello Yashika!
Are you referring to the sum of all 4 digit numbers that can be formed be rearranging the digits of 2468? I will work it out here. If it is some other question, please reply to this comment and I will solve it.
2468 is a 4-digit number in which all 4 digits are distinct.
Step 1: The number of ways in which these 4 digits can be rearranged is 4! = 24 ways.
Step 2: Each of the 4 digits is equally likely to appear in the units place, tens place, hundreds place and the thousands place.
Step 3: So, how many times will 2 appear in the units place.
Of the 24 different 4-digit numbers, 2 will appear in the units place in 24/4 = 6 numbers.
Same goes for the other 3 digits. 4 will appear in the units place in 6 numbers. 6 will appear in the units place in 6 numbers and 8 will appear in the units place in 6 numbers.
Step 4: Sum of the digits in the units place = 6*2 + 6*4 + 6*6 + 6*8 = 6(2 + 4 + 6 + 8) = 120
Step 5: Sum of the digits in the tens place will also be 120. The value will be 1200 as these digits occupy the tens place.
Sum of the digits in the hundreds place will also be 120. The value will be 12,000 as these digits occupy the hundreds place.
And the sum of the digits in the thousands place will also be 120. The value will be 120,000 as these digits occupy the thousands place.
Step 6: The required sum = 120,000 + 12,000 + 1200 + 120 = 133,320
Please reply to this post if you require any clarification in the above explanation.
Cheers
@@Wizako Thank you very much sir, I understand it now, also for repetitive digit question,will Method be same, except the factorial r number of rearrangement, It will be factorial r by factorial x number of ways?
@@yashika0511 Absolutely right. Please try the 2248 question and post your answer.
@@Wizako Hello sir Thank you very much, I have solved 2nd question and answer is 53,328, following are the steps:
1. The number of ways 2248 can be arranged is 4! /2! =12
2. Out of 12, 4 digit number, each of likely to present in unit place 12/4= 3
3.sum of digits in unit place = 2*3+2*3+4*3+8*3 = 48
4. Same sum of the digits in tens place = 48*10= 480
5.final sum of digits after computing hundred and thousands place = 48+480+4800+48000 =53328
@@yashika0511 Absolutely right.
Do problems where order doesnt matter and with replacement also exist? And what is the formula for that?
They do exist. However, they are not tested in the GMAT or in any other major competitive exams. There is no formula to compute them but the brute force method of listing them down.
No one has taught me Stats like this, thank you Sir
Hello Farina
Thanks for your feedback.
You are amazing. Istg words are just not enough anymore. I watch your videos whenever I am stressed about GMAT and you just manage to put me back on track with your amazing explanation of concepts. Thank you so much x 1/0
Thanks for your feedback. Glad the videos are helping you. Best wishes.
Thank u do much sir for ur concepts are the best and easy to grab 👍
The answers for the try at home questions:
1. the sum of rearrangements of 2468: 33330
2. the sum of rearrangements of 2268: 31108
are these correct?
Q1: The correct answer is 133,320.
Method:
Step 1: Compute total number of ways of reordering the 4 digit number 2468.
Step 2: Compute how many of those numbers will end in 2, end in 4, end in 6 and end in 8.
Step 3: Compute the sum of the unit digit of all the numbers.
Step 4: Use that result to find sum of tens, hundreds and thousands place by multiplying the result with a 10, 100, and 1000 respectively.
Step 5 : Add all these values to get the answer.
Do not hesitate to post a comment if you are facing issues getting to the answer.
Q2: The correct answer is 53,328
Method is akin to the one for Q1. Two modifications to be made to the method before computing answer.
1. Total number of rearrangements will be lesser because one of the digits repeats.
2. So, the number of 4 digit numbers that will end with 2 will be twice the number of 4 digit numbers that end with 4 and 8.
Factor these two inputs and compute the answer.
If you end up with an answer other than 53,328 please do not hesitate to post here. I will post detailed solution.
@@Wizako followed your explanation. Got the right answer.
@@Wizako Sir, any shortcut method to find step 2 , here each no will be on unit place for six times. So if any big number is given like 6-7 digit number then how to compute that it will be repeated these many times on unit place?
@@Wizako For 2268 it will be 59994,for 2248 it will be 53328
The explanation couldn't be any better!
Good
🙌
Q2: ans is 53328.
Total combinations possible are 4!/2! = 12. In the 12 combinations each unique digit is repeated 3 times at a specific place and digit '2' is repeated 6 times at each place. The sum for unit's place comes as 48. Similarly, adding other places we get ( 48000 + 4800 + 480 + 48) = 53328.
thanks for the video , sir it really helped
Thanks for your feedback Mehul. The playlist for Permutation has some interesting practice questions. Watch it at ruclips.net/p/PL3JsYBL14ZltoiZSpm5LCDjlYLj7r7k24
Thank you so much sir. All your sessions are extremely helpful for my preparation :)
You're welcome Harika. You can sign up for free @learn.wizako.com to watch our free trial content!
Best wishes for your GMAT preparation!!!
Wonderful class Sir ji
So glad to have found this channel. You are one of the best math teacher! Thank you!
Happy to help Jasmine! Best wishes for your GMAT prep!
your sessions are very informative. Thank you for sharing such an informative session
It's our pleasure in helping you crack the GMAT!!!
I am not able to get correct answer for problem 2 2248, can anyone please help me
Hello Richa, Can you please share how you got to the answer, so that we can guide you better. Best wishes for your GMAT Preparation!
@@Wizako I got the correct answer
I appreciate the videos. You helped me big time in understanding these concepts.
Happy to help you crack the GMAT!!!
Regards,
Wizako GMAT Prep.
Sir, is the total no. of digits repeated is equal to the factorial of its previous digit? like for 3 digits is 2 times each , for 4 digits its 6 times each? is it so on and so forth for 5, 6 digits and so on?
Hello Raisa
Yes - provided all digits are distinct. Else, it will not work.
Cheers
@@Wizako thank you very much , Sir
sir can you plzzz guide with 2248 question ?/??
Hello Shashwat
2248 is a 4 digit number in which 2 appears twice. The number of ways of rearranging these 4 digits is 4!/2! = 12.
2 is twice as likely to appear in the units or tens or hundreds or thousands place as 4 or 8. Let us say, 4 and 8, each appear x times. 2 will appear 2x times. Together, we have x + x + 2x = 12 or x = 3.
So, 4 will appear thrice in the units place, 8 will appear thrice in the units place, and 2 will appear 6 times in the units place.
Sum of the units place = (3*4 + 3*8 + 6*2) = 48
The sum of the digits that appear in the tens place will also be 48 and its value will be 480.
The sum of the digits that appear in the hundreds place will also be 48 and its value will be 4800.
The sum of the digits that appear in the thousands place will also be 48 and its value will be 48000.
The sum of all 4 digit numbers that can be formed by rearranging digits of 2248 is 48 + 480 + 4800 + 48000 = 53,328
Cheers
thank you so much sir.
15:45 if the password is 'YYY' then order does not matter. How to avoid confusion then?
@Kushagra G
Whether order matters comes into picture only when reordering is possible. YYY can be reordered in only 1 way. So, never check whether order matters with a set that has all unique elements.
Check to see whether XYZ is the same as ZYX. If it is order does not matter. If it is not the same, order matters.
Really great!
Thanks again for your feedback. Not sure if you know this.
December is 'data sufficiency' month. Everything that we do on our YT channel is DS related. Please check community tab of the channel where new questions get posted. Video explanations are posted a day or two later. A DS challenge is currently underway. I invite to participate in it. You can click here to get to the community tab ruclips.net/user/wizakocommunity
You're a legend!
thank you for the afterthoughts and the analogy
You are welcome. Sign up as a trial user @learn.wizako.com to access a few other free topics in our portal.
Best wishes for your GMAT Preparation.
sir how shall we use official guide for quant
Hello Qurious
The official guide is the only collection of real GMAT questions (retired ones though). For the quant section, use the official guide to round up your preparation. The OG does not present questions topicwise. So, learn using an alternative source. Practice using an alternative source. That way you will focus on one topic at a time. After completing all the topics, use OG to revise and review and get acquainted with the way GMAT words questions.
As far as verbal goes, you can either follow the same strategy as mentioned for quant or you can mark questions based on the topic area it tests.
For instance, in CR you can mark assumption questions and solve all of them in one go to recap your learning of assumption questions from another source.
You could try Wizako's online course at wzko.in/pro Sign up as a trial user. Try the free topics. If you find the teaching methodology meeting your learning requirements, you can pay and unlock the modules behind the paywall.
Cheers
Thank u very much for this excellent lecture !
Hello Shruti
Thanks for your feedback. Not sure whether you watched all 4 videos in the getting started series. In case, you have not watched all of them, please watch them @ ruclips.net/p/PL3JsYBL14ZltnwdqUt7NTyx9wUzVh-OZu
Best wishes for your GMAT prep
1:14:30
42:04 it should'nt be 100C1 instead of 100 C4 ?
Hi Atul,
The formula is nCr = nCn-r
So for 103C99, it can also be expressed as 103C(103-99) = 103C4.
Both will give the same answer.
If you still have any more doubts, do post it below. Best wishes!
@@Wizako oh I got it now. Sir your lectures are so so so so good. Please make more videos of quants. I really liked you and your concepts... ❤️
@@shemolin_dude Thanks a lot for your feedback! Follow and subscribe to our channel for regular updates. We will be posting more and more videos in the coming days. Stay tuned!
Happy learning!
I am having trouble figuring out the modifications made in Q2. Can you help?
Hello!
Are you referring to this question
In how many ways can a member for the sports committee and a member for the music club be selected from a group of 20 students?
The question does not state that the two posts have to be awarded to two different students. The question permits the possibility of the same student holding both posts. So, it is a case of sampling with replacement.
The posts are two distinct posts. So, in the event of two different students being selected for these two posts, who occupies which post will make a difference. So, order matters.
For example, if A and B are elected to these posts, A sports, B Music is different from B sports and A Music.
If this is not the question you were referring to, could you please post the question here?
Hello! thank you for the reply.
I am referring to the “practice question: Numbers and Digits #2” : What is the sum of all 4-digit positive integers that can be formed by rearranging the digits of 2248. I understand we have two identical digits and therefore some changes must be done to the process used to solve it. What is the method to tackle problems of this kind with repetitions like that?
Thank you very much again!
Step 1: Let us find out in how many ways can 2248 be reordered. It is a 4-digit number with 2 of the digits being the same. So, we can rearrange it in 4!/2! = 12 ways.
So, we will have to add 12 4-digit numbers to find the answer.
Step 2: Each of the 4 digits (consider the two 2s as 2 separate digits) will appear equal number of times in the units place, tens place and so on.
i.e., of the 12 numbers, 3 will have 2 in the units place, 3 will have the second 2 in the units place, 3 will have 4 in the units place and 3 will have 8 in the units place.
So, 6 of the 12 numbers will have 2 in the units place, 3 will have 4 in the units place and 3 will have 8 in the units place.
So, the sum of the units place of all 12 numbers = 6*2 + 3*4 + 3* 8 = 48
Sum of the tens place of all 12 numbers = 48 * 10 = 480
Sum of the hundred place of all 12 numbers = 48 * 100 = 4800
And sum of the thousands place of all 12 numbers = 48 * 1000 = 48000.
Add all of these to get the final answer.
I am planning to shoot a video for this variant. Will upload it sometime soon.
Wizako Thank you very much, the videos are great!
This was so helpful. Thanks a lot
Glad to hear that! Best wishes for your GMAT preparation.
Great explanation sir!!
Thanks and welcome
Hi sir, for the question regarding GRAPHITE where order shouldn't change. As the order matters it shoud be Npr right. Can you please explain this?
Also the standard framework is only for rearranging without replacement right??
If the question had not placed the condition that the order should not change, it would have been a npr question.
However, because you are not allowed to reorder the vowels, the only way you can write the vowels in any 3 out of the 8 places is 'a' followed by 'i' followed by 'e'
@@Wizako Dear Sir, you have mentioned that if order matters then npr . As you wrote above order should not change (not allowed to reorder the vowels) , does not it mean that order matters ?
Great as always !
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@wizako
Hello sir,
I hope you are doing well and thank you for the lecture, sir.
I have a doubt:
In second question of "Rearrangement of letters" (1:09:35), it was mentioned 8C3 *1, as order of AIE should remain the same.
So, my question is that here the order matters, then why are we using 8C3- I mean, for order matters, we use 8P3 (as mentioned in the chart)?
We would have gone with 8P3 provided the question did not place a restriction on keeping the order of AIE unchanged.
8P3 would have re arranged AIE as AIE, AEI, EIA, EAI, IAE, and IEA. But we can only have one of that i.e., AIE.
That is why we did not multiply 8C3 with 3!, the different ways of reordering AIE.
I hope I have answered your query. Please get back by replying to this comment if you still have a query about this concept.
Excuse me sir, the table at 38:40 ...If order doesn't matter and there can actually be a replacement, what is the correct formula?
Is it simply n x r like usual?
Thanks for the videos, really helpful!
It is not tested in the GMAT.
Having said that, the way to compute the number of outcomes for sampling with replacement where the order does not matter is (n + r - 1)Cr
thank you for such clarifying videos. i am working with these to get through gmat. thanks
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