Permutations and Combinations 12 (Sum of digits as integral solution concept) | CAT 2024
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- เผยแพร่เมื่อ 28 ก.ย. 2024
- Permutations and Combinations by Ravi Prakash | Quantitative Aptitude for CAT 2024
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In question “ how many nos upto 1lakh will give sum of digits as 13 it should be 17c4 - 7c4*5
you are pro man:)
Yesss , mine too
yes, it should be C(7,4) but anyway C(7,4)=C(7,3)
Its the same thing. He literally said it doesn't matter if its 7C4 or 7C3, because it yields the same value
both are equal u can take 7c4 also
18:00 min of this video sir please check 7C4 it should be that you are subtracting
7c3 and 7c4 are essentially the same
A+B+C+D=14 and if A=10 or B,C,D any two digit no it becomes 5 digit no greater than 10000 then exclude them
Same doubt
Really great videos sir, love your way of thinking and the way you open my mind! Thank you
Sir, at 18:00 the answer is wrong i hope so waha par 7C4 hoga.
So what it doesnt matter the value is the same💀😭
Ncr=Ncn-r
Thank you sir!!
in last question, can we solve by -
total cases of upto 5 digit - total cases of upto 4 digits = cases of only 5 digits.
im getting same answer as yours sirs.
what if ths sum is a big number, say 20-100 range. then what number should be alott small a
sir in the first question A+B+C+D if sum of digits is 8 then how 1+2+6+0 case be taken it gives sum of digits as 9?
that was just a small mistake
In ques where 1lakh will give sum of 13 ? 7c3 kyu h ?? It should be 7c4 na kyuki 4 plus sign h sir only told us about this nd here he has written 7c3 whyy??
PLzz reply anyone who knows it
@@aditisrivastavaas3177 7c4=7c3 so anyone of it can be written
Thank you @arpita
Thank you Sir 🙏.
Thanks
Taking digits as 14.....us ques mn A>=10 kyu h? It should be A>10 or A>=9 right? Please replyy
a>=10 is there because you can't count 10,11,12...and so on. these numbers are subtracted. so it should be a>=10 or a>9. don't worry, I think you are confusing meaning of inequality signs a little bit. try to understand them on TH-cam. it'll make things clear✌️
Sir, in the first question why have you taken till 4 digits only?
Because you have to find numbers upto 10000 which means 1 to 10000 bus 10000 sum is (1+0+0+0+0= 1) therefore we don't have to include 10000 except that we have 1 to 9999 in which we have 1,2,3 and 4 digit number only
Thank you so much :)
thank u sir
Thank you sir
Why can't 13C4 Numbers (5 Digit Numbers) give a sum of 9? As, 9
bcos 5th digit cant be zero ,if it becomes 0 then it will be 4 digit number example ( 05430 = 5430 ). I hope i cleared your doubt
@@nareshramesh3519 then do (13c4-12c3) in which 13c4 includes all cases up to 5 digits and 12c3 will include all cases up to 4digits their difference will give you your answer.
Lovn' p&c
I attended the classes of Bansal sir in 11&12 th class but believe me sir you are a gem👍
Feels like Magic tricks being revealed! Permutation and combination never felt so easy! Thank You Sir!
At 10:00 how can A get 10 ? Bcoz that will be 2 digit na if a = 10 , then abcd will be 5 digit no , that is violates qus
That's the point, we are removing all the invalid cases.
10, value is 10 , its not that we are putting A as 10 as a digit beside BCD
Keep up the good work. The classes are too good and I Love the sequence how the classes go. A small bit of info- your website is down which would reduce your traffic so please take care. Thanks and cheers!
Thanks
A+B+D+C=13. abcd are digits0to9. Then10,11,12,13, are not under0 to9 we should A+B+C+D=4. 7C3 is answer ...........please explain me....not explained my doubt well please understand
for the sum of digit is 14 case, in one case if we say A >= 10, wouldnt it give all the cases where B will be greater than 10, etc? since this will give us cases in which every shuffling is possible, like sir said in the last sum? so we should not multiply 4 * C(7,3), but just C(17,3) - C(7,3)
In the sum of digits 14 question what if..
A>10 and B>10 also don't we have to count for those cases as well?
and if there's one case where A>10 AND B>10 as well ...according to this method this case will be treated as 2 different cases which is actually a similar case. we removed it twice.
Help, please.
best channel ever
Sir teaches live on Unacademy plus. You can directly contact him there. His full content is not here.
sir,how can we distribute distinct digits using this method as we distribute only similar things using this method?
Think about each digit x as being x number of chocolates.
Number of 7 digit numbers the sum of whose digits is 61 is?
I am unable to solve as the answer is coming in negative. What am i doing wrong?
awesome lecture
i didnt understand the first question
How many 5 digit numbers will give digital sum as 26
tq
;
What will be the answer if the last 2nd question is changed to 22, i.e. How many nos upto 1L will give sum of digits as 22.
Answer
Total cases - 26C4 (14950)
Less:
Cases where either of the 5 digits is more than or equal to 10 = 16C4*5 (9100)
Add: (duplication case)
Cases where 2 digits out of 5 digits is more than or equal to 10 = 6C4*5C2 (150)
Ans: 14950-9100+150 = 6000
Can we get quizz of this lecture?
Do you teach in Time Bangalore?
He is teaching at IMS, Koramangala... Eve 6 to 8 weekday batch!!
@@periyasamyn4857 Koramangala guys are so lucky.
in the first question, total numbers are asked so wont it be (11C3 - 1) as 0 is not considered while counting?
and same will go for every question you solved, as in last question the specification is provided by you only that 0 will be not considered while numbering.
Take these questions as chocolate and its distribution like
8 chocolates should be distributed among 4 friends in such a way that 0 value can be allowed so if we give 0 choclate to 3 friends then remaining one directly comes comes out as 8 "(0,0,0,0) can't be possible" like A+B+C+D = 8 for whole number solution N+R-1 C R-1 , 8+4-1 C 4-1 = 11C3
@@harshitabhansali3638 Yes got it! Thanks )
Sir first question can be done by using AP... An AP with common difference of 9 have same digital sum... As I started it with 8,17,26........9998. There are 1,111 number... Anyone please tell me where am I wrong?
The 1st que is asking about sum of digits not the digital sum so your method is incorrect.
*Can someone tell me if i just dumbly done the same thing long way, or is ir another way of thinkng?*
I tried to solve it with my approach. Finally reached on it but too complex and maybe mine is same and just doing the long way.
Its like : number number can be above 9 so conisder each a case where ever number is 10,11,12,13,14 (or a to say reversly a case where sum of other 3 digits is such that, it causing the 4 th number to go beyond 9).
It was like ,
For every digit 14 turn by turn , (4 ways)+
Every digit 13 , (4 * 3c2)+
Every digit 12, (4*4c2)+
Every dig 11, (4* 5c2)+
Every dig 10, (4* 6c2) =k
So 17c3- k =0 .
*Can someone tell me if i just dumbly done the same thing long way, or is ir another way of thinkng?*
When u learn a good concept, note it. Always utilise the brain in how to make a longer method shorter, never do vice versa :)
Sure . Thanks !
@@Rodha Epic reply sir, still cudos to dhruv, thats a different way to think for sure