Permutations and Combinations 11 (Similar to Different Distribution with atleast atmost concept)
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- เผยแพร่เมื่อ 28 ก.ย. 2024
- Permutations and Combinations by Ravi Prakash | Quantitative Aptitude for CAT 2024
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This guy deserves to be appreciated more.. Has covered everything under one roof.
yes but not everything
@@lornemalvo2171 whatdidheleft
@@lornemalvo2171 🙄
The last question can be done by using the same approach:
A + B + C + D = 10, where A > B which implies A can be of A = B + K form, where K >= 1 ;
on substituting A = B + K in the equation we get
B + K + B + C + D = 10
2B + C + D + K = 10
where B, C & D are >= 0 and K > 1
further simplification
2B + C + D + k = 9
now we have B, C, D & k >=0
making cases now we will get :
11C2 + 9C2 + 7C2 + 5C2 + 3C2 = 125 (Answer)
Did the same way
@@akashuniyal3726 but for the case to be 11C2, doesn't A have to be 0? But how can A be 0 if B has to be less than A?
@@darkdreamer7387 That is why he took k >= 1. Now, A = B + k. Therefore minimum value of A is 1. It cannot be 0.
Thank you Bro ,tgis last Q was a bit tough fot me u made it easy for me Thanks a Lot Bro ... ❤❤❤
can also be done in the method used in previous questions
A + B + C + D = 10
now B can only take values from 0,1,2,3,4 (if it takes 5 or more, A cannot be greater than B as only 10 chocolates are present)
for B = 0, A is atleast 1, hence C + D
I haven't seen a teacher, putting such efforts to make students understand the concepts, and that too for free, even paid classes can't do what you're doing sir, HUGE RESPECT to YOU and Your Methods of teaching. Bhut saare bacchon ki duaayen milengi aapko.
I am really blessed to have Ravi sir for my cat prep, this is my last topic left which will also be completed soon just because of his beautiful teaching method.
Last question can also be done in the method used in previous questions
A + B + C + D = 10
now B can only take values from 0,1,2,3,4 (if it takes 5 or more, A cannot be greater than B as only 10 chocolates are present)
for B = 0, A is atleast 1, hence C + D
👏
I've never seen classes so beautifully explained. Thanks a lot
Sir teaches live on Unacademy plus. You can directly contact him there. His full content is not here.
@@ankitsinha5382 Oh. Thank you. I will
Best best best..... Content and explanation and hardwork and everything .....
Huge respect for you Sir❤
Salute to you 🙏
Providing such best content at free of cost ❤
the best math teacher i have ever seen
SIR THANK YOU VERY MUCH FOR SUCH STRUCTURED CONCEPT NOTES
sir your hardwork shines brighter than anything else,
The last question can be done like this too:
A+B+C+D=10 where A>B
We can logically think here that A will never be 0( as if we take B=0 then A=1)
Case 1: B=0 therefore A>= 1 so a+C+D = 9. So>>>> 11c2. (Notice A min case= 1 we take to RHS becomes -1. Similar to what sir taught in previous questions)
Case 2: B=1 therefore A>= 2 so a+C+D = 7. So>>>> 9c2
Case 3: B=2 therefore A>= 3 so a+C+D = 5. So>>>> 7c2
Case 4: B=3 therefore A>= 4 so a+C+D = 3. So>>>> 5c2
And finally
Case 5: B=4 therefore A>= 5 so a+C+D = 1. So>>>> 3c2
We cannot go further as then RHS becomes -ve.
Just add 11c2+9c2+7c2+5c2+3c2
Add all we get , 125 cases
Sir I think you missed uploading Lecture 9
So that we have to go to unacademy
Thank you so much for letting us know what real teaching is❤💯🙏 always grateful
really beautiful explaination and teaching
This is peak concept clarity
2:57 why 24C4 ?
beautifully structured
Well Structured, appreciate the efforts!.
sir in multiple of question 3 why we this method why do not just do that by old way i.e A+B+C=15 C can be written in the form of the 3k 15-3= 12, therefore 12+2 C 2 = 14*13/2 =91
someone please reply
Very nice content thank you so much🎉
Just one word: YOU ARE GREAT♥
Thankyou sir
Is there any other way of doing the chocltaes and C in multiple of 3 ?
sir i cant find that vedio of ranking
In the last question can we put A=k+1 and B=k then solve for C+D...? Plz, rly sir...
here you are saying A is greater by B by only 1 which is not completely true, A can also be k+2..3..4.....9, iknow im late :P
You can put A= B+K+1 and then you have 2B + C + D +K =9. Now you can put values of B as 0,1,2,..... and find the ans.
no you can't put A=K+1 but you can put A=B+K , where K can have a minimum value of 1 and can have maximum value of 9 and hence you can get a equation of the form 2B+K+C+D = 9.
Hope you UNDERSTAND
Thank you sir
Why in first ques
20 + 4 = 24
This concept of getting 4 assuming abcde are zeroes
I did not get it
And why not 25C5
we have to count '+' sign and not variables.There are 4 '+' signs thats why (20+4)C4
BTW thank you sir!!
hello sir, while solving the last question in this lecture for the solutions of 2A+C+D=10 , u have taken the case of A=0, which is not possible as it is mentioned in the question that A>B . please reply back
Yeah...same doubt
No of cases where A>B = (Total no of cases - Cases where A=B)/2 . He had already calculated total no cases which is equal to 13C3. So he needs to count the cases in which A=B.( thats why he has written A+B+C+D = 2A+C+D as A=B). So at that moment he is not counting the cases where A>B but rather the cases where A=B
And if you feel confused by sir's method you can put A= B+K+1 and then you have 2B + C + D +K =9. Now you can put values of B as 0,1,2,..... and find the ans.
@@aaketchaurasia651 yeah I also initially tried to approach this ques in this way only .........bt got 35 as answer bt d ans is 125. I thought by doing it this way we wouldn't need to make any cases n all as we r directly making d condition(A >B) satisfied by putting (A = B + 1 + k)
If u got my point thn please tell where m I making a mistake
@@mansisharma767 Putting B=0 we have C+D+K = 9 so no of solution = 11C2 similarly for B=1,2,3,4 we have 9C2, 7C2, 5C2 and 3C2. Sum of all these =55 + 36 + 21 +10 + 3 = 125.
Can i get pdf of this solution from anywhere
Thodi mehnat to khud krle bhai
Arey bhai bhai , paper bhi likhdo bhai ka koi jake (from anywhere) 😂
@@SahilRamola
Are bhai tu hi likde likh sake to
@@prasantsharma3935 bhai tera baap chor ke gaya tha ki teri maa ? (with due respect) lol
19:26 yeah, we wouldn't had miss anything if you didn't delete some of the videos
Check unacademy.
@@aswiniskamath8177 that's the whole point? Isn't it?
If i wanted to pay for it, i wouldn't be here in the first place. It's misleading to say "100% course covered" if it's not really the case.
True
I believe he's uploaded all the missing classes, except class 9. But your comment is 8 months old so you've probably already finished this topic :D
He did not delete lecture 9 video. In one of previous video’s comments section he cleared that some of the videos are not labelled sequentially but they all are uploaded sequentially.
This is top level teaching. Thanks a lot.
Thank you so much for these amazing lectures, Referring them for UPSC Csat. 😇😇
i don't know whether you will read this comment or not . but what you are doing cant be expressed in words, the sheer dedication , simplicity and structure all under the same roof. If you launch your brand in the market like a actual coaching like classrooms and material you will be second to none period.
Bhai tune notes bnaya hai kya
@@ADITYASINGH-sw1ns yes
@@mr.shashank3817 bhai share kr skta hai kya exam 8 ko hai to time nae hai abhi
Damn.this is gold mine for ipmat aspirants. calculating number of integer or natural number solutions using this method gonna be so easy!!!!!Thank you ravi sir
15:44 why did sir took c=0 as C=3*0=0 ( which is not a multiple of 3)
I don't think this case is needed 🤔
After watching speed math videos , It's quite easy calculating the values in p&c multiplications.
if he was my math teacher in school i would be in a whole different place by this point of time but its never too late i guess
can't find words to thank you and youtube
At 2:32 how did u get a+b+c+d +e = 20 it should be 30 right because we minus 10 (2+2+2+2++2)from 40
HAPPY TEACHER'S DAY, SIR.
In how many ways 20pens can be given to A B C D E such that A
i got 945 as answer
great teaching, i usually don't comment but you are a lit 🤩
But you commented on 2023 on Lecture 8 of % & Profit loss video
If a denotes the number of permutations of x + 2 things taking all at a time, b the number of permutations of x things taking 11 at a time and c the number of permutations of x - 11 things taking all at a time such that a = 182be, then the value of x is
Can you please explain this
C Always gets in multiples of 3 should not have 0. It should start with 3 onwards
Nice
I have watched distribution part after that I have made Coding ques on codeforces 😁
I dont get the logic behind dividing by 2 in the last question,can someone explain?
sir your p and c part 9 is not available
Thank you Sir 🙏.
Thankyou Sir.
Thank you sir 🙏
Thank you sir
Very beautifully and sequentially indeed sir. :)
Awesome concepts and way of teaching.
The anwer should be C(24;5) when solving as '2a+1'
Tq sir ❤️
Sir, lecture 9 missing.
Shut up
For last question what if A>B>C?
Sir , where can I get your quiz ?
FB group---> Rodha Cat Preparation
He uploads questions there
@@chiragchaudhari6188 hi in frst question if we are taking the even as 2a , 2b etc then also a,b,c,de will be >= 1 . Right?
@@dhruvsharma9083 no it's >=0 bcz 0 is also a even