Permutations Combinations 4 | CAT Exam Preparation 2024 | Quantitative Aptitude
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- Опубликовано: 27 авг 2024
- Permutations and Combinations by Ravi Prakash | Quantitative Aptitude for CAT 2024
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![15 Kdramas SO AMAZING They Make Everything Else Seem Like Trash! [Ft HappySqueak]](/img/1.gif)
Unbelievable speedy calculations from sir 6^3 *7 instantly
Bonus : Find the no of ways to order 10 people such that A>B>C
possible cases = ABDC, ADBC, DABC = 3
total arrangements of ABCD =4!
So no of ways = 3 * 10! / 4!
@@aaketchaurasia651 thanks aaket
Answer would be same... Get the combination of 4 out of 10 number .
So , it would be 6!* 10C4
@@aaketchaurasia651 aaker bhai answer explain karo na nahi samjaaa
@@sonukumar-bd5np how ? why not consider d's rank?
Sir your P&C sessions are THE BEST.
Need to revise this lecture again.
I HAVE A DOUBT SO IN 2ND SCENERIO WHY THE ANS WAS 6 AND NOT 3! WHILE FOR THE SAME TYPE OF QUS A>B>C>D, THE ANS WAS 4!
You can write 3! as well as 6.
Cos 3!=6.
Sir in Q6 (b) - u took 5C2, because colours can't be repeated. But in Q6 (c) - u are multiplying 7 with 6 cube, implying that colours can be repeated? What's the logic?
I think it would have been better if there was a condition mentioned in Q6 (b) i.e. repetition of paints are allowed.
I am also here with the same doubt
Sir please explain this @rodha
The logic lies in the question itself, in the (c) question it is asked that no 2 same colours are adjacent implying that in this particular question the colours can be repeated
ie the colours cannot be the same adjacently however they can be the same alternatively.
In the 2nd question sir assumed the same rules of question (a) which is no repetition of colours as not much information was provided.
Repetition is allowed but no two should be adjacent that's the question...
if you don't understand the concept watch the whole video first it took me till the end of video to grasp what is going on this concept
Sir in question 6 c part, you have used the concept of colour repeating , but why not in other two . I m lil weak in PNC so i didnt understand why if V&B are taken and other 2 cant be Double G or Double R???
same doubt
I don't think 9x8x7 will be correct because we can't choose 1 and 2 at 100th place otherwise the condition H>T>O can't be fulfilled because If we choose 1 or 2 then one of the no.s either in 10th or 1th place will be greater then the no. in 100th place. Same goes for 10th nd 1th place no.s cant be 9 nd 8 respectively because then one of the no.s will be greater then the no. On 100th place . I think it should be 7x6x5/3!=35
That's why it is said that they can be selected in only one way suppose you have a combination of 1,2,3 now you will be putting 100th place as 3 (not 1 and 2)!
in q5 , if we need only 1 way out of 84 ways where h>t>u then why are we just multiplying it with 1 ???? why aren't we dividing?
We don't need 1 out of 84! There is 84 groups, and they cannot be shuffled, so only one way to keep them! 84 groups * 1 way!
Question 5 we can also take zero at the units position so cases will be 84 Plus 9C2 with zero at the unit position
it is in the que from 1 to 9
While writing all the numbers from 700 to 1000, how many numbers occur in which the first digit is greatest than the second digit, and the second digit is greater than the third digit ? 61 (b) 64 (c) 78 (d) 85
Using video's method shouldnt the answer be 3x9x8/6=36 because h can take only 3 numbers (7,8,9) and then t and o can take 9 and 8 and then divided by 3!. but the answer is 85. What am I doing wrong?
Edit: I thought about it. If we make cases like 7_ _ = 7 x6/ 2 = 21 then 8 _ _ = 8x7/ 2=28 then 9_ _ = 9x8/2=36 adding all three we get 21+ 28+ 36 =85.
In Q5 should we not include 0 while calculating, i.e., use 10C3 instead of 9C3 because by using 9C3 we rule out the possibility of having 210, or 320, or any 3 digit number which contains a zero for the matter?
In the question it is mentioned digits from 1 to 9..so 9c3 is correct
It clearly shows 3 digit
Not digit 3
That's the difference
@@edufly129 with digits 1-9 in the latter part of the question
same doubt. Can't use 0 in the hundred's digit though, but what about ten's digit and one's digit?
in case it was 0 to 9.
@26:06 I think it should be 5C2* 4! + 5C1* 4!/2 because we can also choose one colour out of remaining 5. T.he qn doesn't say anything about the colours being distinct
Shreya Chawla no u can't choose only 1 because it will make only 3 clrs ( v & b already n 1 u choose ) if u think u would repeat the CLR then it would be 5c1*4! ( Because only one CLR is repeated ) hope u got it
@@behindthescene4406
Shreya is correct.......
Condition is not specific about number of colors should be der and if we take one color that is being repeated then it ll be 5C1*4!/2 (divide by 2) bcz same colr ll repeat atmost 2 times.
@Sheya Chawla
you are partially right, but I think the question should be more specific. It also doesn't specify that the remaining colors cannot be violet and blue, so cases like (2 V, 1B), (1V,2B) and (2V, 2B) can also be taken
If the question is based on repeating colours then answer is (5 C 1 * 5 C 1) * 4 ! , if not what is the answer I'm getting mad at this Q6(b). Please explain...
Colour will always be distinct
22:46 cutest sir "phol kalals" 🗣
Find the no of ways to order 10 people such that A>B>C
Hey, did you find a solution. I did but it's a bit lengthy
In Question 5, what if H≥T≥O. (explained in replies)
There will be three cases
1) all digits are different - 9c3
2) Two digits are same - 9c2×2 ( the two digits choosen can be written only as 'aab' or 'abb' , a>b)
3) all three digits are same - 9c1 ( any digit choosen can be written as 'aaa' only)
So total ways- 9c3 + 9c2×2 + 9c1
In question 6 (b) why not 5P2 x 4! ?
In the last qn second part, it is not mentioned that the next 2 colorx(except violet and blue) could not be same, unlike the other qns where it is mentioned?
can anyone help me with it?
Last question part 3 is only concerned with 2 colours not being adjacent. Its not related to part 2 of the question where no 2 colours can be repeated lol. So colours can be repeated in part 3 not an issue. Only they shouldn't be side to side(Ariana Grande style haha)
Wonderful classes sir!
i facing trouble understanding question 5.. will watch again next day with fresh mind
Sir 6th Question is harder to understand since Question doesn't tell anything about repeating or distinct colours and in solution did you also include repeating colours?
yes the question should be more precise.
Excellent videp
Thank you sir
Very helpful.
Can someone help me out, at 24:15 7C4 × 4! = 35 × 24
24 is 4!
But how does 35 came please help
7C4= 7!÷(4!×3!) from formula, solve it it's giving 35..... That × 24 is written.....
in question 5) why aren't we considering 9c3 x 1 x 6! ?
Sir please reply.
bcz we have to make a 3 digit no., unlike the previous ques. where we need to arrange 10 ppl according to their ranks(remaining 6 can be arranged in order)
Total different questions previous sums se
lol I’m I the only one who isn’t understanding any pnc videos from rodha ?
In q-5 how can we approach this ques through Method 1 ? Coz if we go by M-1
Thn total ways - 9*9*9 = 729 & divided it by 27 (i.e - 3*3*3 cases ) coz every 1 out of these 27 is our area of interest, gives d ans 27 only .
M confused .......if someone got my point thn please help
Total Ways will be- 9x8x7 (As same number can't be at U, T, H place). And three numbers can be arranged in 3! ways. So, answer = 9x8x7/3!.
@@sudhanshuranjan4862 I don't think 9x8x7 will be correct because we can't choose 1 and 2 at 100th place otherwise the condition H>T>O can't be fulfilled because If we choose 1 or 2 then one of the no.s either in 10th or 1th place will be greater then the no. in 100th place. Same goes for 10th nd 1th place no.s cant be 8 nd 9 respectively because then one of the no.s will be greater then the no. On 100th place . I think it should be 7x6x5/3!=35
@@lostinlife4338 No it should be 7*8*9/3!
Sir in Q5) Why are we not considering 0(zero) in units place..... why are we considering numbers from 1-9 ? why not 0 ?
Because in questions the range is given in between 1 to 9 and not 0 to 9.
Ravi sir is the best. Ravi sir ki jay. East or west, Ravi sir is the best.
Sir , at 28:36 , tha last part should 35*1512.
from every 2 cases we take the case where A>B so why are we dividing 10! by 2?
Because in all the shuffling cases either A>B or B>A thus half of them are cases where B>A out of 10! Therefore dividing by 2 we get the answer
@@ayushkumar-lr3bv
Bro but B > A can't be possible
@@rishipatel4282go through the process once again you'll find it
Fab
Sir Q2 samajh nhi aya
In 5th question,if we take 0-9 digits, then the answer 10c3 ways
No, bcz in Hundreds place how can you take 0 it'll become a two digit no then😂 but question states we need to make 3 digit numbers..... In that case H will only have 9 options (1,2,3,.....8,9) , T 9 options (from 0 to 9 except the preceding one) and U will have 8 options from 0 to 9 but excluding the 2 numbers already taken by HT....
P.s- But I'm not sure about the calculation part.😢😢
@@hritiksha_deb its been so many days, i forgot everything 😂
For 6 th question why do you take 4!❓
Because in question mention that "4 vartical strips"
Thank you Sir 🙏.
@26:05,What is wrong with the thinking that the 3rd box would have 5 options to choose from and the last box would have 4 options to choose from and then these possibilities could shuffle with the given 2 to give 4! and this leads to the answer 20*4! and not 10*4!.I know it is wrong ,but couldnt find the reason.Could anyone please help?
yeah, i think it should be 5p2(permutation) instead of 5c2 since the order of colors is imp(considered by sir in 1st case) making the answer 480.The question isn't well defined. @Rodha
Sir I am unable to understand the shuffling part for A>B>C>D if we chose three possible ways for arranging 3 numbers that's fine but how are we arranging the rest 7 numbers since D might come in between B or C ya phir A And D please help! Understanding this concept
7C4 means u have taken all possible cases of 4 places. In those 4 places I can put A,B,C,D in only 1 way
Does anyone have another way for the 6th ques, part 2? (5C2 × 4!)
4C2×2factorial(choosing 2 spots out of 4 for V and B and choosen spots can be shuffled in 2 factorial ways)×5C2 ×2factorial(choosing two colours out of remaining 5 and choosen two colours can be shuffled in 2factorial ways)....4C2×2 factorial×5C2×2 factorial
(5x4)x2!x3!
@@mohdsaifkhan3974 after 5×4 why can't it be 4!for arranging all 4
At 23:22 I think the answer should be 7C4*4!/2 . Because in 4! Combinations colour will repeat from left and right. Like VBGO and OGBV are the same thing.
Then even VGBO & VOBG & BVGO & BGVO &BOGV all are the same thing.
That's why he multiplied with 4! For each combination of colours. The question didn't say that the colours can't be repeated
Hey can be repeated that's why he put 4! Or else he could just put 7C4 and left. Those are the number of combinations possible
But he put 4! Because repetitions are allowed
@@kritikasingh7992 it is of permutations the so why dont go by permutation only thAT IS 7*6*5*4
@@muskanchhugani7272 it is not of permutations
@@muskanchhugani7272 Even this gives answer of 840 only na.