Sorry if this was already mentioned but in 30:14 there is a small mistake: it's not "smaller than 4000 i.e. up to 3999" but "not greater than 4000 i.e. up to 4000". So the correct answer is 3x5x5x5 + 1 (which is "4000"). So the answer 376.
an excellent video which i wish would have been there when i was appearing my 10th boards. Express my heartfelt gratitude pushpinder and perfect scores for having made the concepts of permutation and combination so easy.
At 4:28 the answer would be 4)625 if the question was opposite ? Like if the children come to get the chocolates instead of chocolates been distributed to them ?
I think the problem at 35:50 is solved wrong. A "Does not want to sit on even no of chairs" -> according to your solution A wants to sit on even number of chairs.
You can solve these questions other way around too, though that may be tedious. for instance, at 23:55, you can assign tasks to people as well i.e people3 gets task2 (1 way), people1 can get any of tasks{3,4,5,6} i.e 4 ways and similarly people2 can get 3 no. of tasks(this is because people1 and people2 cannot get task 1). Now we have 3 tasks and 3 ways i.e 3!. which equates to 4*3*1*4!=72. Similarly, if people4 gets task2 instead of people3 we get 72. Adding two we can 144. It's true that this method is more confusing but i just wanted to establish that we can solve questions other way around too. As an exercise, Ive solved every question other way around!
You're actually doing a great job by providing tutorials on something so easily accessible as RUclips. It's helping a great deal. Thanks for that! But I'd like to bring your attention to the glitches only so you can improve the quality of your videos. There are quite a few silly mistakes in the narration and notations that leave me confused momentarily. If you could have scripts or shortforms to read while you record, it would make you much more efficient. Also, the "not greater than 4000" condition was wrongly interpreted.
see as two conditions are mixed together so we have to take different cases and while taking cases we have to consider the mixed case only once so B having 2 places to sit in 1st case sits on one place and in second case sits on the 8th place
Hey Pushpinder, you active here!? Wanted to point out a flaw in one of your questions' answer. "But not greater than 4000" takes 4000 as well. Answer, hence, must be everything that you said i.e. 375 + 1 = 376.
The answer for the problem at 31:00 is not 375, but 376, because you also need to add 1 option where the number is 4000 (not greater than 4000 means that it can be equal to 4000)
@PerfectScoresThank you for this video. I have been struggling with this topic all my life. Now it is clear. PLEASE could you tell me where I can find practice problems?
The question says,Integers NOT GREATER THAN 4000 i.e 4001,4002....so on are not included and hence 4000 is included. It nowhere says that less than 4000.
Great Video...Very Useful... But I would like to point out that at 31:45 for the prob on Integers the answer would be 376 because 4000 also has to be added since in the ques they have asked >>not greater than 4000.
at 44:19 in the second case if we give 2,4 to the 4th slot then shouldn't we be having 2 choices again for slot 1 since 0 cannot be filled there and we have to make a 4 digit number?
Correct, that's what I thought as well. We can have only 2 choices for the 1st Slot as 0 cannot be considered for the 1st slot since it's a 4 digit number.
There are no negative cases here. One case: A boy wants to sit in only the even numbered chair, and second case: another boy wants to sit at only the extremes.
Thanks, Video was very useful, perfect explanation! Please help! how can i use the same approach for the following question? Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?
+Anna Abzianidze 6!-5!*2! Where 6! is the total number of possibilities and 5!*2! is the chance that Marcia and Jan are together. To get this possibility, you consider Marcia and Jan as 1 item, so you get a total of 5! of them being together. But, you also multiply by 2! because inside the "item", there are 2! total possibilities. Once you have that, yous substract this number to the total of possibilities
At 24:32, the CAT question; the negative condition is task 1 cannot be assigned to EITHER person 1 OR to person 2, so wouldn't it be FOUR ways you can assign task 1 instead of THREE that you said in the video?
At 31:37, we can still have 4000 since the question says the number can not be more than 4000, which takes the total number of ways to 376. Am I right or wrong?
Hi ankit, as the video was taking cases , as we are not giving any choice to B sit on any of the chair , we asked her to sit only on 2nd chair in fist case and 3rd chair on the 2 case , as per the condition she wants to sit on any of them , when we add it , she only had two choices , and we distributed in 2 cases.
ankit rathore As much I understand by rechecking the concept in the starting questions, when you are using non colliding conditions, you are supposed to take both cases i.e 2*3*6! but when you have questions with colliding situations, we will take one position at a time so 1*3*6!.
At 31: 51, the answer should be 376 coz the question says not greater than 4000 meaning less than or EQUAL to 4000 and hence adding 1 i.e the number 4000, the final answer is 376. Excellent video though. Thoroughly enjoyed it!
For last question when you look at the scenario where 4 th place is given either to 2 or 4 you say that 1st place has 3 choices left meaning you include 0 but 0 cannot be included because then it wouldn't be 4 digit number.
No, he has not included 0. because 4 place can have only one Digit that is either 2 or 4,but not both at a time, if we use 2in last position "4" will be left with us ie first place can be filled with (1,4,5)
No, he has made a mistake as he has counted 0 as well for the 1st position. The reason being, he considered 0,3 and 5 as 3 options for the 1st slot. So there's a mistake here.
I think there is a problem at 44:10 as when we take 2,4 as options for the fourth digit we are left with 0,1,5 for the first digit place but if we take 0 it won't be 4 digit number please help me out with this as you took 3 instead of 2 here
mistakes in 34:30,sir in dis given question particular boy doesnt want to seat on even number of chairs but u fixed them on even number of chairs.......so please check n correct it.....
either way the answer won't change...coz the other boy wants only extremes and the arrangement will cause you to get the same answer..although better to stick to the qn since this is not always gonna happen.
PerfectScores Vishal Jamwal If u use that logic then at time 37:13 the answer should be(2*3*(6!))+(2*4*(6!)) but the video shows (1*3*(6!))+(1*4*(6!)).
At 4:32, can I consider each child can get five chocolates (Every child can have any number of chocolates) at most and by the procedure explained before can this be right ->5*5*5*5 ?
The answer should be 4*4*4*4*4 = 1024, because the chocolates are to be distributed, so question solved according to chocolates rather than children. 1st chocolate has 4 choices, 2nd also 4 and so all five have 4 choices of children to get distributed.
problem at 11:00. shouldn't it be that girl A has 4 chances (2nd, 3rd, 4th, 5th) Girl B would then have 3 chances (5 prizes minus 1st place and the one A got leaves 3) and the last 3 have 3 prices left so 3! for them should equal 4 X 3 x 3! = 72. you got the same answer but a way i don't understand.
Hi Pushpinder, I could only see Permutations & combinations I and II. For III and more it asks to subscribe. however at the end of your video II, u have mentioned that they will be free. please provide access to those as well. thanks
Sir in the part 36:00 in the 2nd case shouldn't B have 2 choices left 1 and 8 and out of that B has chosen 8 because of which A has only 3 options left? Please explain! Apart from that thanks a lot sir for your efforts.. really appreciate it! :)
it's a awesome video.....concept is very clear.yes,if we go in a structured manner its easy....thank u..plz reply to my question...i'am not able to solve that.
in the collision case, in your examples, +ve case wont be having 2 choices either of only one choice? because question mentions that the +ve case have 2 choice
It was a very helpful video. However I have a doubt. What if the question has both positive conditions or bothnegative conditions. How to resolve such questions, please let me know.
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Sorry if this was already mentioned but in 30:14 there is a small mistake: it's not "smaller than 4000 i.e. up to 3999" but "not greater than 4000 i.e. up to 4000". So the correct answer is 3x5x5x5 + 1 (which is "4000"). So the answer 376.
You are the best explainer I have found on youtube!
Dear teacher, I am from Iran and found your videos really helpful. Thank you so much.
an excellent video which i wish would have been there when i was appearing my 10th boards. Express my heartfelt gratitude pushpinder and perfect scores for having made the concepts of permutation and combination so easy.
At 4:28 the answer would be 4)625 if the question was opposite ? Like if the children come to get the chocolates instead of chocolates been distributed to them ?
I'm pausing mid video and just wanna say thank you for making this, i've been struggling with this topic my whole life and it helps me a lot!
Do yourself a favour and watch this video at 2x. The best video on P&C on the internet!
I was struggling in this bcs I forgot the concept. U just made easy. Thanks.
answer at 31:44 should be 376 because 4000 will also be included as a 4 digit number which is not greater than 4000.
how??? can you explain me? i didnt get it
The answer is 376. Annotation should be correcting that.
ayush chauhan Isn't the answer 500? (4*5*5*5)
Yup exactly .. coz the questions mentions not greater than 4000.. so we add the 1 condition of 4000 to 375 .. making it 375 + 1 = 376
exactly one case for including 4000
Superb, I didn't get in the classroom but here you have explained very structured and proper way. Thank you buddy
ay 29:39 for the CAT 2008 que, it says not greater than 4000, so it can be equal to 4000, right?
Yes it should be 376.
I think the problem at 35:50 is solved wrong. A "Does not want to sit on even no of chairs" -> according to your solution A wants to sit on even number of chairs.
true that, but still the answer is going to be same!
yea i he have considered even numbers where in A does not wants to sit on even numbers
Yeah you are true he made a mistake
@@jaychavan653 answer wouldn' be same bro
I got 7! + 6×6!
though there are some mistakes but what an awesome way of teaching I was facing a lot of difficulty with this u made it easy for me thanks alot sir
Awesome lecture Mr.Gill, Iam satisfied with ur explanation and numerous examples.. I appreciate ur effort.....god bless u
This video truly brought me out of the bombardment of P&C . thanx alot pushpinder.
Very nice examples from ground root level. Perfect for understanding PnC !
Thank u so so much..i really was struggling wit this topic for quite sometime now ..you made it remarkably easy to understand ...!!
Great video getting concepts of permutation and combination thoroughly cleared. Thanks sir
you are the best teacher in the world !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
You can solve these questions other way around too, though that may be tedious. for instance, at 23:55, you can assign tasks to people as well i.e people3 gets task2 (1 way), people1 can get any of tasks{3,4,5,6} i.e 4 ways and similarly people2 can get 3 no. of tasks(this is because people1 and people2 cannot get task 1). Now we have 3 tasks and 3 ways i.e 3!. which equates to 4*3*1*4!=72. Similarly, if people4 gets task2 instead of people3 we get 72. Adding two we can 144. It's true that this method is more confusing but i just wanted to establish that we can solve questions other way around too.
As an exercise, Ive solved every question other way around!
at 30:20 greater then 4000 means doesn't it mean that 4000 also includes in the range??
Ya based off that language 4000 should have been included imo
And would be 376 in that case
376 is answer
Thank you so much. This is a great video. It's so much better than my textbook.
A very good tutorial to learn basic of Permutation and Combination arrangement cases.
You're actually doing a great job by providing tutorials on something so easily accessible as RUclips. It's helping a great deal. Thanks for that! But I'd like to bring your attention to the glitches only so you can improve the quality of your videos. There are quite a few silly mistakes in the narration and notations that leave me confused momentarily. If you could have scripts or shortforms to read while you record, it would make you much more efficient. Also, the "not greater than 4000" condition was wrongly interpreted.
Yes he missed one case where number can be 4000
I love this tutorial and your fashion sense.
until now i was struggling in the topic, this did help me a lot. Thanks a ton!
As Russell Peters would say FANTASTIC!!!!!
it's really nice explanations with lot of examples ...thank you sir
At 36:15 doesnt B have two ways , as B can either choose 1 or 8. I think it must be 2*3*6! . I am confused. Please make it clear for me.
see as two conditions are mixed together so we have to take different cases and while taking cases we have to consider the mixed case only once so B having 2 places to sit in 1st case sits on one place and in second case sits on the 8th place
@@Vyankateshpatole she must be 4 yrs older by then.... and lol u are one year older while i am replying
thanks a lot for presenting the "conditions topic", interesting and very useful
its way to do:
conditions first ,solving next
Hey Pushpinder, you active here!? Wanted to point out a flaw in one of your questions' answer.
"But not greater than 4000" takes 4000 as well. Answer, hence, must be everything that you said i.e. 375 + 1 = 376.
Thank you for the feedback.
The answer for the problem at 31:00 is not 375, but 376, because you also need to add 1 option where the number is 4000 (not greater than 4000 means that it can be equal to 4000)
Good job Pushpinder sir.
All the videos are totally helpful in all cases.
Thank you for the interest.
This is extremely helpful and well-explained. Thanks very much!
@PerfectScoresThank you for this video. I have been struggling with this topic all my life. Now it is clear. PLEASE could you tell me where I can find practice problems?
Thank you very much. You can access all our content at perfect-scores.com
I really loved the way you have explained. great work and outstanding tutoring.
Thank you.
Thanks a lot for this video.
Also, you might want to consider using a PC pen and pad instead of writing with a mouse.
Used rn it was 8 years ago, but for you at your old time it was 4 years ago.. Then how do you expect him to use ipad
Thank you pushpinder sir. Your voice is amazing. Little soft.
at 31:12 the question says less than 4000 so 4000 cant be a number but in the annotation says 4000 is also included
The question says,Integers NOT GREATER THAN 4000 i.e 4001,4002....so on are not included and hence 4000 is included. It nowhere says that less than 4000.
man tysm for this video and i will make sure to check ur other vdso aswell since i am preparing for gmat this feb 1st,hope i do well :) thanks again!
Great Video...Very Useful...
But I would like to point out that at 31:45 for the prob on Integers the answer would be 376 because 4000 also has to be added since in the ques they have asked >>not greater than 4000.
Yes
Superb video.
one comment : the problem of making numbers >999 to not >4000 the solution should include 4000 also according to question.
sorry i see the correction already in the video
Best video for permutation and combination concept
at 44:19 in the second case if we give 2,4 to the 4th slot then shouldn't we be having 2 choices again for slot 1 since 0 cannot be filled there and we have to make a 4 digit number?
Correct, that's what I thought as well. We can have only 2 choices for the 1st Slot as 0 cannot be considered for the 1st slot since it's a 4 digit number.
nope since the place can be filled in two but will eventually take only one digit leaving the other no. for filling the first place
31:30.. it is mentioned "not greater than 4000". that means we have to include 4000 also.. so the answer should be 376..
how??? can you explain me? i didnt get it
He calculated the number of terms from 1000 to 3999.. but question says not greater than 4000.. so we're supposed to include 4000 here.. so 375+1
+Rishank Devadiga ok got it...thanks
you're preparing for which exam?
+Rishank Devadiga gre
Great work. I like how many examples you did. Thanks for making this video.
though there are some mistakes the video is very helpful .. i liked it very much .
at 36:54 the question says A does NOT want to sit on even numbered chair.
Does it not mean that A wants to on odd numbered chair?
There are no negative cases here. One case: A boy wants to sit in only the even numbered chair, and second case: another boy wants to sit at only the extremes.
love 8920505536
Thanks, Video was very useful, perfect explanation! Please help! how can i use the same approach for the following question? Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?
+Anna Abzianidze 6!-5!*2! Where 6! is the total number of possibilities and 5!*2! is the chance that Marcia and Jan are together. To get this possibility, you consider Marcia and Jan as 1 item, so you get a total of 5! of them being together. But, you also multiply by 2! because inside the "item", there are 2! total possibilities. Once you have that, yous substract this number to the total of possibilities
in the question at 36:51 u considered the case of boy siting in even chair which contradicts the question
Dude ur a really good teacher...keep up the good work👍🏻
Thank you Sir for the great tutorial. Excellent for concept building.
31:45 the answer is 376 not 375 as case of 4000 will also be included.
At 24:32, the CAT question; the negative condition is task 1 cannot be assigned to EITHER person 1 OR to person 2, so wouldn't it be FOUR ways you can assign task 1 instead of THREE that you said in the video?
At 31:37, we can still have 4000 since the question says the number can not be more than 4000, which takes the total number of ways to 376. Am I right or wrong?
@29:14 regarding the question with 2 cases.
B has 2 ways to sit on the chair as B can sit on 2 or 3 but u gave it 1.plz explain
Hi ankit, as the video was taking cases , as we are not giving any choice to B sit on any of the chair , we asked her to sit only on 2nd chair in fist case and 3rd chair on the 2 case , as per the condition she wants to sit on any of them , when we add it , she only had two choices , and we distributed in 2 cases.
sachin dargude
I am sorry, the questions was taking cases not the video
ankit rathore As much I understand by rechecking the concept in the starting questions, when you are using non colliding conditions, you are supposed to take both cases i.e 2*3*6! but when you have questions with colliding situations, we will take one position at a time so 1*3*6!.
Sanket Saxena yeah even I have the same doubt becaz ans. Will be different then
Great video
Also, at 31:45 it is mentioned "Not greater than" so number 4000 is valid, so total ways must be 375+1=376 ways
I believe the same thing.
376 is the correct choice.
At 31: 51, the answer should be 376 coz the question says not greater than 4000 meaning less than or EQUAL to 4000 and hence adding 1 i.e the number 4000, the final answer is 376. Excellent video though. Thoroughly enjoyed it!
For last question when you look at the scenario where 4 th place is given either to 2 or 4 you say that 1st place has 3 choices left meaning you include 0 but 0 cannot be included because then it wouldn't be 4 digit number.
No, he has not included 0. because 4 place can have only one Digit that is either 2 or 4,but not both at a time, if we use 2in last position "4" will be left with us ie first place can be filled with (1,4,5)
No, he has made a mistake as he has counted 0 as well for the 1st position. The reason being, he considered 0,3 and 5 as 3 options for the 1st slot. So there's a mistake here.
Thats a great video. I am feeling very confident in this topic. Sir, Thank you very much.
I think there is a problem at 44:10 as when we take 2,4 as options for the fourth digit we are left with 0,1,5 for the first digit place but if we take 0 it won't be 4 digit number please help me out with this as you took 3 instead of 2 here
At 31:10 answer should be 376, question says an integer not greater than 4000, so integer can be 4000 as well.
I thought that your answer is wrong because not greater than means less than right
I thought that your answer is wrong because in the question they mentioned not greater than 4000 which means less than 4000
Such that 4000 wouldn't include
Great job! Now I finally got the concept
Nice and advanced examples! I just finished a short Combinations explainer video on my channel.
At 44:42 sir if you give 2,4 to last then you cant give zero to ist so you are left with 2 choices ..then how 3
dude ur voice is very catchy . great work
+Rashmi Pandey No, its not. Its rather annoying.
no its not so
At 30:10 shouldn’t it be
@36:35 - I'm confused - if he wants to sit at extremes then shouldn't 1 and 8 be the extremes?! meaning 2nd case: 2 x 3 x 6! ? - Thanks in advance.
this is what even i got!!
love 8920505536rej
2nd case would be 1X3X6! because we have already assigned the 8th position to B so B has only 1 way.
Mira Guru so is it 2x3x6! (when B sits on 1st extreme) + 2x4x6! (when B sits on 8th place)
very nice explanation
Thanks a lot for such comprehensive explanation
mistakes in 34:30,sir in dis given question particular boy doesnt want to seat on even number of chairs but u fixed them on even number of chairs.......so please check n correct it.....
either way the answer won't change...coz the other boy wants only extremes and the arrangement will cause you to get the same answer..although better to stick to the qn since this is not always gonna happen.
at 36:35 u said wrong sir.the boy does not want to in even chair.
Rest was really helpful.Thank u !!!
thanks alot ...best video ever..plz upload more of it
You can get it here:
www.wiziq.com/course/69618-perfect-scores-premium-tv-gre-gmat-cat
PerfectScores Vishal Jamwal If u use that logic then at time 37:13 the answer should be(2*3*(6!))+(2*4*(6!)) but the video shows (1*3*(6!))+(1*4*(6!)).
At 4:32, can I consider each child can get five chocolates (Every child can have any number of chocolates) at most and by the procedure explained before can this be right ->5*5*5*5 ?
ya it should be 5*5*5*5
not 4*4*4*4*4
because each child can can get in 5 ways
i.e, 1st child 5ways
2nd child 5ways
3rd child 5 ways
4th child 5 ways
The answer should be 4*4*4*4*4 = 1024, because the chocolates are to be distributed, so question solved according to chocolates rather than children. 1st chocolate has 4 choices, 2nd also 4 and so all five have 4 choices of children to get distributed.
problem at 11:00. shouldn't it be that girl A has 4 chances (2nd, 3rd, 4th, 5th) Girl B would then have 3 chances (5 prizes minus 1st place and the one A got leaves 3) and the last 3 have 3 prices left so 3! for them should equal 4 X 3 x 3! = 72. you got the same answer but a way i don't understand.
very useful.I understood clearly.thank you
Hi Pushpinder, I could only see Permutations & combinations I and II.
For III and more it asks to subscribe. however at the end of your video II, u have mentioned that they will be free. please provide access to those as well. thanks
Sir in the part 36:00 in the 2nd case shouldn't B have 2 choices left 1 and 8 and out of that B has chosen 8 because of which A has only 3 options left? Please explain!
Apart from that thanks a lot sir for your efforts.. really appreciate it! :)
First of all thank so much for your video Sir :)
Waiting for your next video on Permutation and Comb..
appreciate that you fixed few errors with annotations :)
could you please explain further how the question at 4:39 you got 4.4.4.4.4.4=1024 ways? I think you fell short of explaining in the video.thanks
KWAKU OSEI-TUTU I think answer might be 625
very structured procedure. thanks
My pleasure
I appreciate your work, very helpful video !
How many four digit numbers can be formed that has the digit 5 used exactly once in them
You really made it so simple..🙏
it's a awesome video.....concept is very clear.yes,if we go in a structured manner its easy....thank u..plz reply to my question...i'am not able to solve that.
love 8920505536
In question 32:02, greater than 999 but not greater than 4000, 4000 should be inclusive and the answer should be 376...correct me if I am wrong
in the collision case, in your examples, +ve case wont be having 2 choices either of only one choice?
because question mentions that the +ve case have 2 choice
at 31:07 u mentioned 4 cannot come but 4 can come as 4000 is allowed? am I right on that?
The answer is 376
why other videos are private?
In the que. where we are supposed to find the 4 digits even number why 0,2,4 are not considered as single entry??why is 0 and 2,4 taken separately??
Awesome concepts,..But the 2nd video on P&C needs more clarity... Anywaz it was very good...Thanks..!!
29:52 its 376 because we can consider 4000
isnt it?
It was a very helpful video. However I have a doubt. What if the question has both positive conditions or bothnegative conditions. How to resolve such questions, please let me know.
Good explanation
And btw sir, THANK YOU! Your class is really interesting!
tq u its really so clear to understand the concept
Thank you very much....it is much simpler now
Thank you for your feedback!
In questions where there are 5 prizes and 5 girls with conditions, if there were instead 4 girls. Would the method still work?