Permutations Combinations 2 | CAT Exam Preparation 2024 | Quantitative Aptitude
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- เผยแพร่เมื่อ 28 ก.ย. 2024
- Permutations and Combinations by Ravi Prakash | Quantitative Aptitude for CAT 2024
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better solution for question 7:
Case 1. The numbers in which 3 occurs only once. This means that 3 is one of the digits and the remaining two digits will be any of the other 9 digits
You have 1*9*9 = 81 such numbers. However, 3 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 3 will appear only once.
Case 2. The numbers in which 3 will appear twice. In these numbers, one of the digits is not 3 and it can be any of the 9 digits.
There will be 9 such numbers. However, this digit which is not 3 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 3 is written twice. Therefore, 3 is written 54 times.
Case 3. The number in which 3 appears thrice - 333 - 1 number. 3 is written thrice in it.
Therefore, the total number of times the digit 3 is written between 1 and 999 is 243 + 54 + 3 = 300
Nice approach.... It helped, thnks!!
Gr8
This was really helpful
in the first case, there will be a situation when 3 will occur at 2nd and 3rd position, in that case, the 1st position will have options from 1to 9 that is 9-1 = 8, that means the number of ways here will be 8*9*1 + 8*1*9= 144+81= 225?
But answer can't be 300 because in case 1 ,2 and 3 888 will repeat.. so answer should be 300-2 = 298
for ques 7, we have to exclude some common numbers from all 3 cases, e.g 333, 330,303,33,etc. Answer will be 289
We needn't. I understand that you are saying this because we might double or triple count them. But if you think about it we need to double and triple count them. 333 will be counted once in all three steps, meaning it will be counted thrice [which is equivalent to the number of 3s it has]. We have to find the number of 3s and not the number of digits that contain at least one 3 which is discussed in Q10.
@@ptmohankumar thanks for this doubt clearing comment :). This thing Was really messing up my mind then I saw this.
@@ptmohankumarwow... thanks brother...i was confused too ... as you said "We have to find the number of 3s and not the number of digits that contain at least one 3 "
@@ptmohankumar excellent explanation brother 👍👍
I needed the answer too..
Tysm 🔥🔥🔥🔥
@@ptmohankumar very nicely explained
You are the most down-to-earth, selfless person to come across youtube. you really have the best of wishes from your students. The way you teach is absolutely inspiring.
Q 10, for people who are confused:
(No. with at least one 6) = (all possible numbers) - (numbers without 6)
All possible nos. b/w 1 and 1000 inclusive is 1000
1-3 digit numbers without 6 = 9*9*9 - 1 (as 9*9*9 includes 000)
total numbers without 6 = 9*9*9 -1 +1 (to include the 4 digit '1000' that was missed in the previous step
(No. with at least one 6) = 1000 - (9*9*9 - 1 + 1) = 271
arree bhai ye +1 kis bt ka h?
@@keshavmeena1161666 ek baar aaraha h woh bhi ek single case haina
@@keshavmeena1161666 ek baar aaraha h woh bhi ek single case haina
Yeah sir made a mistake here
@@keshavmeena1161 +1 is for the number "1000" which wasn't included in previous case as he was counting 1-3 digit number only and 1000 is a 4 digit number
Q 4, 9 * 10 * 10 * 9 (overall choosing) / 10 * 9 (changing values) which will give non changing values of palindromes.
in question 10 it says "at least" not "atmost" so should we consider all 10?
Best teacher for aptitude exams on earth 🌎 ❤
to everyone facing difficulty - there are 271 nos b/w 1 to 1000 which has 3 in it but 300 times 3 appears in those 271 numbers. Cheers!
271 will be the total numbers having 3 in them, but here the question is the number of times it has appeared. For instance 333 will be a number that is in your list of 271 numbers, but here 3 coming 3 times in ‘333’.
Thanks
And how to get this 271
In question 10 for those who is facing difficulty in understanding,
_ _ _ = (10x10x1) 100 ways
0-9 0-9 6
_ _ _ =(10x1x9) 90 ways
0-9 6 0-9 (excluding 6 as we cannot repeat)
_ _ _ = (1x9x9) 81way
6 0-9 0-9
Excluding 6 Excluding 6
Total= 100+90+81 = 271
This is another way.... it is correct
I found this one to be the simplest and quickest :)
the solution for how many times 3 will appear is first we find all values where 3 comes single time which is 300- 30(where digits are repeated twice or thrice-033,133,233,333,433,533,633,733,833,933,303,313,323,333,343,353,363,373,383,393,330,331,332,333,334,335,336,337,338,339). Now we know these 30 terms have a total of 21*3 3's present in them but we have taken 333 3 times so subtract 6 from the final tally of 63 digits. We get 57 times the digit 3 appears. Now, we have to add the total 57 to 270 giving the final answer as 327.
If you have any doubt in question 7 it would be clear in 10th.
In question no. 10, to calculate total no. , why we didin't use formula = diffrence +1. As by that way tota no should be = (1000 - 1) + 1 = 1000
I did the same thing.
The queston says in how many numbers 6 appears atleast once and i number 1000 you can clearly see no 6 is present , so he removed it and took total 999.
@@siddharthyadav9625 but is it neccessary to neglect 1000 during calculation...
Because it seems neccessary to neglect 1000
As answer will get changed if we consider 1000 ..... i.e, 1000-728=272..answer is not right..
So it is mandatory to neglect 1000 which is bit strange
Sir you teach very well👌👌
For Q 7, you can solve the question check the answer on excel. I have done it, sir's answer is correct.
In Q7 If times is asked then the Answer will be 300. Since 1 to 100 ,20 times 3 is coming (11 from 30 zone and 9 from rest no.s like 3,13..etc) Then multiply it by 9 we get 180. Now for 300 to 400 there are 100 + 20 extra from normal 11+ 9 way. And 333 is already taken into counting since we have taken 33 in 2 times and 300 to 400 case 100 times. So sir way is absolutely correct. Hope my ans will clear the concept
James bond of quant 🙌007
In "Number of 3 in between 1 to 1000" question,, from CASE 1: There is possibility to get 033, Same for CASE 2: we can get 033
So 33 is repeated and we are not excluding it from final answer, why???
Yes u are right I have the same concern
Because question is 'how many times 3 will appear' and in no. 33 , 3 is appearing twice ,so it has to be added twice and similarly 333 can also be repeated in all 3 cases but is not excluded since 333 contains 3 3s that needs to be added three times
3 is repeated twice in 33
questions 7 to 10 are having ambiguities.
Yeah, u r saying right , I also didn't understand
Sir. If we include 1000 then also we need to take 000 out. So Answer will be 272. So why didn't included 1000. Because we have to count for 1-1000 . 900 also not included 6 so y we had taken out that.
But sir in question 10 why we not counted 1-1000... Please clear then also 000 will not occur & our answer will change...
In question 5 can't we have a palindrome of ABBBA ? If so then won't we have to add 5*10 = 50 ?
but 3-3-3 will going to include in all three cases then we will going to get 333 three times so why we are counting like this?
A huge respect to you sir🙏🙏...you deserve all the happiness
in 4th question, we also consider aaaa possibility for palindromic numbers?
Abba + aaaa will sum up to same total.. hence it was included
Thank you sir, Q 10 was the best of all. Made all the concepts crystal clear .
16:55 what about the case of 333 being counted thrice ??
Sir what is the difference between how many numbers have 6....and numbers with atleast one 6...if any numbers has 6 we calculate all numbers with 6 and if it has atleast 1 "6" then also we consider all numbers with 6 so why ....ans is different 300 and 271 ??????
For question 4 it can be all a and all b option also na?
In the second method of last question... Sir n bola k 999 digits we will take qki 1000 m 6 include ni h but 6 to ase bhot saare no.s m inc ni h...
Guys who are confused in Q7 , try counting digit 3 between 0 to 100 ,using above method. And count by normal method also. then you will understand.
😂
🟠⚪ Great fruitful csat session ⚪🟠
in question 7 aren't we counting same numbers in different cases like 333 can be possible in case 1,2 and 3.
doubt with ques 7...i think the first case will take care of all possiblilities
Thank You Sir! Understood the topic like never before!!
thank you sir
Sir thanks a lot. 😍😍 Your way of teaching is amazing 🥳
In Q7 at 14:55, when we add all the cases, 333 will be counted thrice. Shouldn't be substract 2 at the end?
That's a very valid point, although 300 is the correct answer. I don't know how it fit...
I figured, when we fix a number we count the digit only once even when it's appearing twice. we count it again when we fix the other digit. Do it from 0-100 you'll understand.
In 333 the number 3 counts as three
Easiest form simply 1000-729=271 ans
.😅
Great job sir 🙏🙏
in the 7th question won't 333 will be repeated three times if we just add all the 100 from the different three position without subtracting the common count i mean i googled the answer its the same but it should be a bit different if we think of this from the multiplication addition approach
The question is asking how many times the no 3 will appear. So for e.g., in no 33, 3 is appearing two times, so it will be counted as 2, for the no 333, there are three 3s, hence it is counted 3 times.
By considering 333 only once, we are just counting one 3, but the question is demanding how many times '3' is appearing, hence thrice.
I hope it helped ☺️
@16:13 , ans. should be 100 because we are repeating all those numbers in case 2 & case 3, which are already in case 1, If I am wrong then correct me.*****
how can we repeat nos which does not include no with digit 3 at unit digit .....Ex 234, 267, 458 etc. These don`t lie in 1st case
In 2 and 3 case 333 is also coming like 1 case
AbbbA is also a palendremic number , but in this case there will be only 250 such numbers
Yeah...But in such cases you have to take the maximum possible number of palendromes
How 250 can you explain??
Sir in paletromic number if a is 9 so b should also be 9 as 1 digit is used
Q 7 if we see logically then 1 to 1000 only 271 times 3 appear then how 300 , whowever made this equation or formula doesnt he or she thought about logic
In 10th que they only ask the no of digit 6 will appear atleast once ? Then the ans is 243 then again why we are adding 28 in which 6is occuring twice or thrice?
How many 3 digit even numbers are there which are distinct
Sir you made10th qstn more complicated.
sir please suggest from where should i practice questions for eaach chapter
Sir in Q7,for example,numbers like 333,you are considering in all 3 cases.This answer is wrong.
In case of a number like 333, we have to add +3 to our counter as the number 3 is appearing thrice in Three Hundred Thirty Three.
So we need to consider it thrice!
No bro ...because in 333 there are three 3's ...so we need to count it three times.
In 4th question, 4 digit palibdromic numbers between what sir ?
sir why didn't you make total no.s available =( 1000-1) +1=1000 in ques 10?
For question no 10 we could've take 1 to 1000 as well right? Because the answer is same when you solve by including 1000
from 1 to 1000 including 1 and 1000 there are 1000-1+1 = numbers(between 1 and 10 there are 10-1+1 numbers including 1 and 10)
@@premsiramant1321 Why did we not include 1000 in case of six? Last question Total numbers will be 1000
Im still confused when to use addition and when to use multiplication :(
In qus 7 i thought we need to do x but the solution was to be found using + menthod.
The best one is here
@16:35 sir numbers like 333 , 313, 332, 233 are included in more than one case
Sir is right.
You are counting such numbers where at least one 3 is present.
Sir question 7 is correct? U make 3 cases... Okay... But in every case has same number like 333 Or other too so please make me clear this
Answer is coming as 300 only with the different approach.
Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.
Case 1. The numbers in which 3 occurs only once. This means that 3 is one of the digits and the remaining two digits will be any of the other 9 digits
You have 1*9*9 = 81 such numbers. However, 3 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 3 will appear only once.
Case 2. The numbers in which 3 will appear twice. In these numbers, one of the digits is not 3 and it can be any of the 9 digits.
There will be 9 such numbers. However, this digit which is not 3 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers.
In each of these 27 numbers, the digit 3 is written twice. Therefore, 3 is written 54 times.
Case 3. The number in which 3 appears thrice - 333 - 1 number. 3 is written thrice in it.
Therefore, the total number of times the digit 3 is written between 1 and 999 is 243 + 54 + 3 = 300
P.S. this is exact copy paste from Beatthegmat.com to save myself from typing
@@utsavs6429 Absolutely, and 333 coming thrice is actually giving the count of 3 3s in 333. Otherwise 333 would have been counted as a single 3.
Sir for ques 7.
Case 1 may produce 033
Case 2 may produce 033
This is also counted twice..
I think case we must do it like..
Case 1 = 10 X 10 X 1 = 100 ways
Case 2 = 10 X 1 X 9 ways(excluding digit 3 at unit's place) = 90 ways.
This removes double counting of 33,133,233,333,433,533,633,733,833,933
Case 3 = 1 X 9 X 9 =81 ways
This removes duplicate counting of 330,331,332, 333, 334, 335,336,337,338,339,303,313,323,343,353,363,373,383,393
So case 1+2+3 = 100+90+81= 271
Is this correct?
in ques 10 its asking atleast isnt this formula of atmost?
hello sir. 4444, 5555 i.e. aaaa arent these type of numbers palendromic?
Hats off !!
I glad to your sir really 👌🙏
sir 333 is counted in all 3 cases .
Kisis ne iss pureyy. Pnc course ka notes banaya hai kya please reply?
26:40 can any 1 explain why shoud multiple 9 with 3?
Bro watch the same part of video
U will get your answer
In ques 10 what's the reason for including those last 2 steps of 666 ---27 and 666---1. Please explain.
Those steps are there to
include numbers who have digit 6 in them twice for e.g 626,636 and those numbers who have digit 6 in them thrice for e.g 666
@@adityamankar8910 Thanks for the explanation, I got it!!
You are brilliant! sir 🌼
Sir.. 1 to 1000..has 1000 nos naa... how did u take 999 nos..!?
At 20:45 won't 8888 repeat in all cases
sir in Q10 can we simply do 1000 - 729
How 999 total numbers in question 10 ?
Sir for ques 7.
Case 1 may produce 033
Case 2 may produce 033
This is also counted twice..
I think case we must do it like..
Case 1 = 10 X 10 X 1 = 100 ways
Case 2 = 10 X 1 X 9 ways(excluding digit 3 at unit's place) = 90 ways.
This removes double counting of 33,133,233,333,433,533,633,733,833,933
Case 3 = 1 X 9 X 9 =81 ways
This removes duplicate counting of 330,331,332, 333, 334, 335,336,337,338,339,303,313,323,343,353,363,373,383,393
So case 1+2+3 = 100+90+81= 271
Is this correct?
Bhai the question is how many times digit 3 will appear.In case 1 we are considering the no. 3 on units place In case 2 we are considering no. 3 on ten's place. When you say no. 33 in case 1 here the 3 of the units place is considered along with 03,13,23, 43 etc... And in case 2 only ten's place 3 of 33 is considered along with 31,32,34 etc
Right.
Double counting would happen if the question would ask about how many numbers are there which have 3 contained.
Good content.
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Confusing
Please put video 9, 20
ig there is mistake in question of problem 10
Awesome
ty sir
15:00 20:00 28:00 32:00
Sir for ques 4 ans must be 81
As 90-9
Or a=9
B=9
9*9 81
Otherwise a and b will be same
Show some more energy while teaching.. Otherwise good content
Question 7 answer is wrong..
Coz it considering some number multiple times, like in first case you can take 333 and for the second case you could choose 333..
So you can find solution by follow
you can do it with negation , like
Total numbes-(number which doesn't contain 6)
Then,
Total numbers = 1000
And
The numbers which doesn't contains 6 which can be find by follow..
Numbers of one digit which is not 6 are = 9 (0 excluded)
Numbers of 2 digits(i.e 10 to 99) which is not contain 6 = 8 x 9= 72
Numbers of 3 digits(i.e 100 to 999)
Which doesn't contain 6 = 8 x 9 x9 = 648
So total number are which doesn't contain 6 = 9 + 72 + 648 = 729
So the numbers which contains 6 are = 1000 - 729 =271
Answer=271
You are wrong here completely. Think over it :)
@@Rodha yaa you were Right, sorry sir.. i got my mistake..
Q10 is confusing
think over it. spend some time
sir, in last question it says from 1-1000 shouldn't we include 1000 too?
no bro as from 1 to 1000 we are doing calculation for 3 places keeping in our mind so if u take 1000 then things will change...and in case u calculate using 1000 numbers then also u have to deduct a single number for 1000 itself as there is no 6 in number 1000...
W
Question 10 mein It should be How many time 6 exist exactly once from 1 to 1000. At least 1 mean > or =1 333 will be counted in at least one but it won't be counted in exactly one.
u are right i have same doubt
For Question 7:
HIS APPROACH IS CORRECT. But it is a little nuanced. Here, he counted the number of possibilities of all the numbers that has 3 in either one of the unit's, ten's or hundred's or two of them, or all of them. Which means that the number 333 has been included thrice and the numbers that contain 3 two times are included twice. The question here asked is the number of times of 3 appears from 1-1000. So the the number that contains 3 thrice has been counted thrice. Same is with the numbers that contain 3 twice. He's GOAT.
Sir, your way of teaching is really applauding... Thank you for selfless service... And helping us
Sir, why we didn't use 10 X 9 in Q7 , as the question says SEQUENTIALLY. I'm not able to understand please help .
sir in question regarding digit 3 appearance in numbers from 1 to 1000 where we have taken 3 cases and then at last did the total of the three cases for the final answer . but sir numbers like 333 ,233 ,133 are repeating multiple times and we are counting them multiple times . so is it what suppose to be done ??
The question asks about the number Of times the digit 3 will appear and not about how many numbers are there containing 3. In case of numbers like 333 , it is repeated in all the three cases because it contains the digit 3, three times. So it has to be counted thrice.
@@46-suvendusamantaray49 Why did we not include 1000 in case of six? Last question
for quest 7.
in case 1 : 3 is at units place, hence we will get a number 333 once.....we will get this same number in case 2 where 3 is placed in middle and we will get the same number 333 when 3 is placed at thousands place in case 3........hence we will be counting 333 thrice. This way we will count the 3 '3's in 333.
Hope it makes sense.
thanks for this clarity
In question 10 why we are not subtracting 1 to 5 which also don't contains 6? So the total will be 9994 instead of 999?
In question 10, It should be, EXACTLY once, ( 6 comes exactly one time), Atleast means >=1. which will be counting number of 6 from 1 to 1000.
Same Doubt with me..
I think it should be atmost 1
exactly
it should be at most once or exactly once.
It should be exactly 1 or else the answer should be 300
sir,In Q10 1 to 999= 270 and as the question is 1 to 1000--As "1000" is not considered in 1 to 999, as No:1000 also donot have any "6"
can we take 270+1(includes no:1000)=271 as answer! is this method correct or wrong? please clarify this sir.
Sir in question 7, Numbers such as 333 and some numbers are counted more the once .How should we do that?
same doubt digit 333 is counted for 3 times..
@@bankingsaga8020 Yes, but we must also remember that 333 has three 3s. So, 333 coming in the 3 cases actually accounts for the three 3s. Think about it, you will get it. You are required to count the number of 3s.
@@sanchayanmondal5068 This was helpful. Thanks.
@@bankingsaga8020 Since the number 333 has three 3's. So in the first case in the number 333 we've counted only 1 three and that's why in the next two cases the number 333 will also repeat for counting another two 3's
@@sanchayanmondal5068 please elaborate a lil more in detail
can't we say we added 1 in question 10 to include 1000
why is 1 not considered when counting total numbers between 1 to 1000 but 0 is included when counting total numbers between 0 to 999?
The way you take concepts from zero to advanced is amazing. Wow, What a Guru!. Thanks a ton.
In question no. 10, the question is "In how many numbers, digit 6 will appear at least once from 1 to 1000?" so, while taking all possible numbers between 1 to 1000 we have not considered 1000 as it won't have 6 in it, so similarly we could have excluded 1 also as it doesn't have a 6 in it. Like, is the question somewhat incomplete? I haven't understood this concept. Can anyone please explain?
Instead we can consider total numbers available from 1 to 1000(including both)- numbers which doesn't contain 6
i.e. 1000- (9*9*9)=271
I'm grappling with the same issue. My advice would be to see the logic behind our operations. We are actually just counting with shortcuts. We are observing cases when it's a single digit no. then when it's a two digit no. and then when it's 3 digit. 1000 is a four digit no. and our question scope isn't there, so we don't need to worry about it. hope it helps.
It almost feels illegal to watch this content for absolutely free ngl❤
In question number 7,won't it count 333 in all cases?Please explain it sir.
Is there pdf available for these lectures?