Practice Questions on "Permutations and Combinations" available at: www.careerride.com/mcq/permutations-and-combinations-quantitative-aptitude-mcq-questions-40.aspx
Hii sir U said in 4 th question like 3×6×5 ,here instead of putting a 4 said 5 chances,but if we place there 0 in third place it implice chance to get a 500,but in question asked to find out between 500 and 1000 means 500 not applicable.
The fact that sir says 'easy' before each sum gives me the encouragement before starting every problem idk why. Thanks a lot for all your effective sessions
For Question 12 at 1:01, it is useful to mention that we can obtain the answer without calculation. It is all the numbers from 1 to 999 plus 000 which will be a 1000.
My goodness. This channel is so great and the explanation is genuinely very easy. For a person who doesn’t get maths, going through all yt channels who just chitchat and blabber, is just tiring. But this channel focuses on explaining the concepts and boosting your confidence so well! I am so grateful I came across this channel.
Sir, I wish to know your good name. The way you say it’s easy, actually created within us a sense of confidence and makes us to work on it even if the question is quite tough to solve. Your effort to teach is reflected in your kind voice tone.
Thank you Sir for such a marvellous and unforgettable explanation. I skipped this chapter in high school around 15 years ago. Now I'm compelled to learn it to achieve a milestone and just went through it once and the whole concept is crystal clear. Thanks again.
What I like best about ur video is variety of questions nd no faltu k reading comment nd responding to faltu ppl who disturb classes. I really liked d video. Value of students time.
Trick 01: 6C2 = 6/2 × 5/1 (Subtract both the numerator and the denominator by 1 until you get the denominator 1. This will make your Combination calculation a lot easier.) More examples; 18C5 = 18/5 × 17/4 × 16/3 × 15/2 × 14/1 Trick 02: 17C9 is same as 17C8 because 17-9 = 8 More examples; 11C3 => 11C8 8C3 => 8C5 18C11 => 18C7
This 1-hour video helped me understand and solve combination & permutation than an entire semester of statistics course! Thumbs up! Wish me luck for my exams!
I really appreciate your teaching style sir.you always solve every questions so easily.I am learning all the arithmetic topics from you to clear my doubts.and now I am able to solve problems. Thank you so much sir.
Sir, I am from Bangladesh. I am pleased to this video and have acquired a clear conception regarding these 15 math solutions. Hope you show more videos in coming days regarding remaining different rules on permutation and combination problems with detail explanation on all theorems. Thanks
You're welcome, Urooj! Do share the video with your friends and family and join us in our mission to make good content available to all. Stay connected :)
Sir, thank you so much I thought it was tough but now it is so easy.. Your way of teaching is so effective and much interesting.. Once again thank you🙏
When I read the question I was like oh my God how to solve this sum.. but when you say , it is easy sum then I get confidence within myself... thank you very much sir. Because of you I m able to solve the sums.😊.
I was always skiping these topics ( permutations and combinations, probability, profit and loss) but after watching this explanation, I felt this topic is not so difficult
you don't know how much this basic and simple thing and the way of explanation has helped a lot in appearing NET exams. I'm a MSc scholar now but keep forgeting this basic formulas and concepts because we don't use these concepts in day-to-day life and just watching this video reminds me of everything I've locked up inside a small space in my brain. thank you so much.
Hey careerride you have not included sums like "how many triangle or straight line can be formed" and variations on these sum. If you could please make a video on those sum as well. As always your channel is GOLD.
Still figuring out what to do with permutation and combination is a headache for me. I mean the question can be understood differently and the number of conditions that we can apply for each question is still a wonder
Great sir...very nice video There is small mistake...when u discus 4 qus...when repition allowed...146 will vome...becz if repition allowed 500 cant come so 146 ways
i have watched all the videos for aptitude and reasoning he says this is easy, very verry easy, extremely easy question, this question seems hard but very easy
I have always neglected Permutations and combinations questions asked in exams because I didn't know how to go about it. It seems easy and tempting to try but without basic concept of it, could never get it right . But from today, especially after watching and learning the basic concept, I m 100% sure and confident to take any questions from permutations and combinations that may be asked in my future exams. All this is possible because of you sir. Again thanks a lot Would love to keep learning from you May god bless you🙏
But when we selct three alphabets and it contains one k..it can be any on of the two k's so we have to eliminate that case When there is one k..because of duplication every possibility repeats so we have to divide with 2 and subract from 8c3
you,ll find your ans after reading definition of combination carefully, it gives no of combinations for selecting r DISTINCT objects from n DISTINCT objects. DISTINCT WORD IS WHERE ANS TO YOUR DOUBT LIES.
Pradhosh - Take a look at QA and Reasoning play lists on our channel. They are pretty exhaustive and have helped a lot of students crack these placement tests. Should be helpful to you too. All the best :)
In ques 13, we have only 9 ways to fill the hundred's place because we have to exclude 0. If 0 comes at the hundred's place then number will be no longer of 3 digits,it will become a two digit number.
For q4 repitation case we would have to do 147 -1 =146 as one number that would be formed would be 500 in that case and we have to choose between 500 and 1000
I think in ques 4 if repetitions are allowed answer should be147-1= 146 because 500 will be included in the answer but we don't require that . if I am wrong correct me. btw series is amazing :)
13th question considering he starts at zero seconds then for 2nd attempts he will take 6 seconds i.e., (2-1) * 6 seconds for 3rd attempts he will take 12 seconds i.e.,(3-1) * 6 seconds therefore for 1000th attempt he will take 999 * 6 seconds = 5994 seconds
Bro your teaching style is very good, Though level of questions is not that tough, but taught me an important lesson, always start with easy problems first then move to hard ones
Sir, can you tell me one thing that at 3:49, why do I have to arrange a thing in a circular table(or format) in (n-1)!, why not in (n!) ways?? please explain sir...I want to understand this one logically, not just by memorising formulas, so please explain me
1:05:30 14) To Choose at least 1 yellow chair from 3 chairs [ 6 - Non-yellow Chairs and 3 - Yellow chairs ] out of a total of 9 Chairs = (3c1 * 6c2) + (3c2 * 6c1) + (3c3 * 6c0) = 45 + 18 + 1 = 64 Ways Thanks Me later.
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
Practice Questions on "Permutations and Combinations" available at: www.careerride.com/mcq/permutations-and-combinations-quantitative-aptitude-mcq-questions-40.aspx
Can u pls explain the last question....why we took 6c³ for the last selection... without j
Thank u soo much sir
Hii sir
U said in 4 th question like
3×6×5 ,here instead of putting a 4 said 5 chances,but if we place there 0 in third place it implice chance to get a 500,but in question asked to find out between 500 and 1000 means 500 not applicable.
Sir, in Qno.9 you repeat 7 two times , whether the Q says no repetition of digit.
@@uppalabhanu1168 repeatation is not allowed and in 500 there are two zeros
The way u say..."how easy it was...?" makes things really very easy....😀
correct✌️
i think you should show this teacher face who educated more than million people on aptitude
Yes
Yes highly agreed
Yup I agree
Yup
does face or name matter?
After all the ups and downs in search of permutations and combinations tutorials, this makes more sense to me, thank you so much.
When he says very easy then actually everything seems to be easy . Love the way you teach sir ❤️
Yes, Prakriti. That's how our mind works. :)
Thanks.
The fact that sir says 'easy' before each sum gives me the encouragement before starting every problem idk why.
Thanks a lot for all your effective sessions
Word !
so true!
For Question 12 at 1:01, it is useful to mention that we can obtain the answer without calculation. It is all the numbers from
1 to 999 plus 000 which will be a 1000.
My goodness. This channel is so great and the explanation is genuinely very easy. For a person who doesn’t get maths, going through all yt channels who just chitchat and blabber, is just tiring. But this channel focuses on explaining the concepts and boosting your confidence so well! I am so grateful I came across this channel.
Long live this man! You saved me in my exam... Thank you so much sir
"See How easy it was" is an emotion ♥️🔥🥀
and should be the way of life :)
Sir, I wish to know your good name.
The way you say it’s easy, actually created within us a sense of confidence and makes us to work on it even if the question is quite tough to solve.
Your effort to teach is reflected in your kind voice tone.
Thank you Sir for such a marvellous and unforgettable explanation. I skipped this chapter in high school around 15 years ago. Now I'm compelled to learn it to achieve a milestone and just went through it once and the whole concept is crystal clear. Thanks again.
What I like best about ur video is variety of questions nd no faltu k reading comment nd responding to faltu ppl who disturb classes. I really liked d video. Value of students time.
Trick 01:
6C2 = 6/2 × 5/1 (Subtract both the numerator and the denominator by 1 until you get the denominator 1. This will make your Combination calculation a lot easier.)
More examples;
18C5 = 18/5 × 17/4 × 16/3 × 15/2 × 14/1
Trick 02:
17C9 is same as 17C8 because 17-9 = 8
More examples;
11C3 => 11C8
8C3 => 8C5
18C11 => 18C7
Where in the video tho ?
@@theorist353 at 1:11:34
bro how to apply the tricks in question suggest any video of permutation & combination
This 1-hour video helped me understand and solve combination & permutation than an entire semester of statistics course! Thumbs up! Wish me luck for my exams!
All the best!
Reply back - 2
it was a very effective session for me , each and every part you have teached here was very fluent , i really appreciate your teaching , loved it
Honestly this is the first time I actually understood P&C. Thnaks alot Sir..
You're welcome, Megha. Stay connected :)
Sir,Your each lecture increase my confidence level to solve Aptitude test 😊
Thank you sooooooooooooooo much! If I clear my exam, it will be all because of you!!! You are seriously way better than my own coaching teachers!!
All the best! Keep us updated.
did u cleared bro!!!!!
I really appreciate your teaching style sir.you always solve every questions so easily.I am learning all the arithmetic topics from you to clear my doubts.and now I am able to solve problems.
Thank you so much sir.
Hyee 😊
When sir says easy, I become calm and I feel its actually easy!!😍😍😍
Regarding 4th problem, if there is repetition then the answer will be 146, bcz we have to exclude 500 since it doesn't lie btn 500 n 1000.....
Yes i also saw that this vid was good but i am searching for a vid only in letterrs pnc
Yeah.. Even I was thinking of the same thing!
No, It would be 147, because 0 is not repeating it will come only once so the number will be either 50X or 5X0 ... here X will be anything except 0.
@@PawanChaudharyRudeBoy Excellent
@@PawanChaudharyRudeBoy But it is ;ossible to fill unit and ten's place in 7 possible ways right?. So 00 is possible
After many time finally I understood P&C. Thank you sir 🙏🙏
Sir, I am from Bangladesh. I am pleased to this video and have acquired a clear conception regarding these 15 math solutions. Hope you show more videos in coming days regarding remaining different rules on permutation and combination problems with detail explanation on all theorems. Thanks
Convert into Hinduism then and be obliged
Anyone from 2024?
Watching
🙋
From 2057🤚😅
Few hours before exam
present
Excellent effort . thanks for clearing my concepts
You're welcome, Urooj! Do share the video with your friends and family and join us in our mission to make good content available to all. Stay connected :)
Sir, thank you so much I thought it was tough but now it is so easy.. Your way of teaching is so effective and much interesting.. Once again thank you🙏
When I read the question I was like oh my God how to solve this sum.. but when you say , it is easy sum then I get confidence within myself... thank you very much sir. Because of you I m able to solve the sums.😊.
creame biscuit
Was very useful...huge respect for you and your team...please make more videos on various topics in quantitative aptitude
I haven't studied permutation combination with that much interest, it is all because of you... thank you careerride
Life becomes easy when you a good teacher like this
Sir your way of teaching is very good.
Really appreciate it.
Thank you ❤️.
I was always skiping these topics ( permutations and combinations, probability, profit and loss) but after watching this explanation, I felt this topic is not so difficult
1.75x is the correct speed to revise these videos
you don't know how much this basic and simple thing and the way of explanation has helped a lot in appearing NET exams.
I'm a MSc scholar now but keep forgeting this basic formulas and concepts because we don't use these concepts in day-to-day life and just watching this video reminds me of everything I've locked up inside a small space in my brain.
thank you so much.
Keep teaching in English, thank you 🙏🏼
Thank u sir....💛❤️ ಧನ್ಯವಾದಗಳು...😊
explaining easy things in a complex way
EXTREMELY HELPFUL! I really really appreciate your time and gift for teaching!
Hey careerride you have not included sums like "how many triangle or straight line can be formed" and variations on these sum.
If you could please make a video on those sum as well.
As always your channel is GOLD.
Thanks for the suggestion, Shivam. Will try. Stay connected :)
The content on this channel is so useful, it really clears my basic 😊
Man just loved the series of logical and aptitude from career ride, thank you so much it helped me a lot for entrances
Still figuring out what to do with permutation and combination is a headache for me. I mean the question can be understood differently and the number of conditions that we can apply for each question is still a wonder
You are the best! You outshine all! Excellent simple explanations.
Great sir...very nice video
There is small mistake...when u discus 4 qus...when repition allowed...146 will vome...becz if repition allowed 500 cant come so 146 ways
Yes
Super explanation sir i never saw any one to explain like this sir
You're amazing. Thank you so much!
i have watched all the videos for aptitude and reasoning
he says this is easy, very verry easy, extremely easy question, this question seems hard but very easy
Yes, Ritik.
That's how our mind works. :)
Thanks for watching the videos.
I have always neglected Permutations and combinations questions asked in exams because I didn't know how to go about it.
It seems easy and tempting to try but without basic concept of it, could never get it right .
But from today, especially after watching and learning the basic concept, I m 100% sure and confident to take any questions from permutations and combinations that may be asked in my future exams.
All this is possible because of you sir.
Again thanks a lot
Would love to keep learning from you
May god bless you🙏
Really the best channel ,thankuuu so much sir
Anyone watching this in October 😵💫2024?
Very nice vdo ...i learned so much that i didn't even expect....precise and just to the point very well explained
What an awesome teaching 👌👌👌👌
‘See how easy it is ‘ is lub.🤗
I understood permutation and combination conceptually only after watching this video...thanks a lot🙂
Me too
really your teaching technique is too much better than other .....
Legend watching a few hours before exam
Thankyou so much 🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
This helped me a lot
In the easiest wayyy😊😊😊😊😊
37:19 in question 5 .... Why we are taking 6C2 and not 6*5 in sub question 1 ... Similarly in sub question 2 6C4
Same doubt @dineshmantri6658
thank you for sharing your tricks with all🙂🙂
in question 15 why cannot we take direct 8C3 as it also contains all combinations including 2K s and 1K s and no Ks.?
Same doubt.. @careerride pls clarify
Me 2
For me too
But when we selct three alphabets and it contains one k..it can be any on of the two k's so we have to eliminate that case
When there is one k..because of duplication every possibility repeats so we have to divide with 2 and subract from 8c3
you,ll find your ans after reading definition of combination carefully, it gives no of combinations for selecting r DISTINCT objects from n DISTINCT objects.
DISTINCT WORD IS WHERE ANS TO YOUR DOUBT LIES.
badhiya tha, sab saamagh agya
In the 4th problem answer is 146 as 500 cant be between 500 and 1000....We have to eliminate one possible arrangement from 147 possible arrangements
the numbers are not supposed to be repeated, in 500 num 0 is repeated so ans 147
is correct
Thank you so much ..ur videos helped me a lot ...I got placed in TCS
How did you prepared for aptitude?
That's a fantastic news, Selva. Congrats and all the best! :)
Pradhosh - Take a look at QA and Reasoning play lists on our channel.
They are pretty exhaustive and have helped a lot of students crack these placement tests. Should be helpful to you too. All the best :)
@@CareerRideOfficial Thank you:)
I like the way you always mention that the question is not difficult !
You see....How dealing with the mind is more important than dealing with the question :). Stay connected
@@CareerRideOfficial awesome bro
Thank you sir
sir on 1:10:02 can we arrage the remaining letters in (8C3)/2! since we just choosing, its logical if "k" is repeating twice?
Ty🙏
Thank you i learn Something.,
sir you should always tell the logic behind everything you explain
best video on youtube for permutation and combination
In ques 13, we have only 9 ways to fill the hundred's place because we have to exclude 0. If 0 comes at the hundred's place then number will be no longer of 3 digits,it will become a two digit number.
but here it is password, we can create a password starting with zero, zero will be considered if we create a password starting with zero
Every vedio is beneficial after 6 years also
Appreciating a lot
Salute hai sir aapko
SIR U R THE BEST! Thanks a lot
For q4 repitation case we would have to do 147 -1 =146 as one number that would be formed would be 500 in that case and we have to choose between
500 and 1000
can you explain
We can't minus in repetition
it's really very helpful... thank you very much... 🖤
Thank you sir for sharing your knowledge with more than a million people
I think in ques 4 if repetitions are allowed answer should be147-1= 146 because 500 will be included in the answer but we don't require that . if I am wrong correct me. btw series is amazing :)
I was about to comment the same.
The question says BETWEEN so we dont include 500 & 1000 🙂
53:54 You could also have solved it directly using a permutation, as travelling from A->B =/= travelling from B->A. Really helpful vid, thank you :)
You are right because the order does matter whether the person is moving from left to right or front right to left ❤
Started Watching your videos recently and it helped me a lot sir, Thank you🙏🏽♥️
teaching approach is very best
Great video thank u soo much, it helped me a lot
Nice video sir.. removed all my confusion
Watch them in 2x speed.(it will save you a lot of time)
Great videos for quick practice.
note - 57:43 imp,1:02:09 10c3 3! 6 ,different meaning ,atleast =all-(~desired) ,q15
Sir in question no. 4 , in case of repetition being allowed i think the answer will be 146 since we need to remove the case where we get 500.
best video of P&C available at youtube👌👍
Thanks, Kamal. Please subscribe for more career related videos and stay connected :)
Thankyou so much, your videos are best explained. Please keep uploading such videos on youtube for everyone
okk dear
what an amazing class.. one hour went like 10 minutes
13th question
considering he starts at zero seconds then for 2nd attempts he will take 6 seconds i.e., (2-1) * 6 seconds
for 3rd attempts he will take 12 seconds i.e.,(3-1) * 6 seconds
therefore for 1000th attempt he will take 999 * 6 seconds = 5994 seconds
UR CRT
If it is a 3 digit lock I don't get how there are 10 digits for first place. Shouldn't it be 9C1 X 10C1 X 10C1 i.e. 9X10X10 = 900 combinations?
@@deepikaniroula7333 passwords can start with any number
011 can also be a lock pswd
Thank you sir so nice of you sir
For 44:03 we can also do
10c1*12c0+10c0*12c1.
Bro your teaching style is very good,
Though level of questions is not that tough, but taught me an important lesson, always start with easy problems first then move to hard ones
Excellent bro! You way of explaining is great. Thanks alot
Bro what r u doing now
Thank you sir ur explaining way is very perfect way.
Sir, can you tell me one thing that at 3:49, why do I have to arrange a thing in a circular table(or format) in (n-1)!, why not in (n!) ways?? please explain sir...I want to understand this one logically, not just by memorising formulas, so please explain me
Yeah sir even I didn't understand it
30:37 the answer with repitition should be 147 -1 =146..
Because the repitition also acccounts for the number '500' , which shouldn't be counted
1:05:30
14) To Choose at least 1 yellow chair from 3 chairs
[ 6 - Non-yellow Chairs and 3 - Yellow chairs ] out of a total of 9 Chairs
= (3c1 * 6c2) + (3c2 * 6c1) + (3c3 * 6c0)
= 45 + 18 + 1 = 64 Ways
Thanks Me later.
Anybody else loves his tone ?