To make it easier, in Ques2, you can simply take a+b+c = 28 (where a, b, c >= 0) hence, we can have cases from 0,0,28 to 14,14,0..making it total 15 unordered cases or 15*3=45 ordered cases. I hope this helps. Cheers!
Similarly, for ques 3, you can take a+b+c = 21 (where a, b, c >= 0) hence, we can have cases from 0,0,21 to 10,10,1...making it total 11 unordered cases, which includes 7,7,7 thus, (11-1)*3=30 ordered cases. I hope this too helps. Cheersagain!
On @@bhargaviagrawal7216 's Note. If you solve by 2A+B method. You would only get 10.5 on keeping C as 0. So we tend to go to 10. But remember A can have 0 . By that method also we will get 11 unordered cases.
@ 10:52 instead of taking the equation as A+B+C=37, we can also take a+b+c=28, where a=b => 2a+b=28, where a can take 15 values from 0 to 14, thus counting 15 values of the digits repetitive case.
19:52 since 2A+C=24 A can take values from 1-11 = 10 values(cannot have 0), and 3A = 24 in this case A=8 we consider it as 1 possiblity but why we are not subracting it from that 10 values since it also have the value 8 in it?
Hi chauhan ,,here we are not distributing the 24marbles to 3 groups ,we are making 24 marbles into 3 groups,so whiling making groups we cant assig 0 ,bec group means atleast 1 ,so hope u understand:)
in question where we distribute 24 identical marble to 3 groups why we need to give atleast 1 to a b c its marble that we are distributing not people so even if group can not have 0 people they can have 0 marbles . if somebody can explain please do.
@@turkishnaxal1703 if you put 24 marbles on a table and distribute them in 3 groups you will need atleast one marble in a group. if there is no marble, the group does not exist at all ,you will see only two groups on the table.
In how many ways can 9 identical balls be distributed among four baskets such that each basket gets at least one ball? what would be the answer for this ?
Absolutely fantastic approach to provide an edge over other aspirants. I have a query sir , If in the question (similar to similar) we have to distribute among more than 3 boxes, bags , groups etc. Then, how will we approach the problem because then while we convert ordered to unordered cases would be more not just (a,a,b) and (a,a,a). Wouldn't it be a chaos if the question is like how many ways 6 identical books can be distributed in 8 identical boxes. If we approach with A+B+C+D+E+F+G+H = 6. It won't be possible to convert 13C7 into unordered.
Sir in the question - No of ways of distributing 37 identical books to 3 identical bags such that each bag contains atleast 2 books, here in (a,b,c) cases there will be some cases where *a* , *b*, *c* will be less then 3 i.e 1, 2, 34 so why you haven't subsracted those cases??
There is some mistake in question... Of 37 books in three bags... The answer( 30C2-3*5)/6 is not integer... How's that possible... There must be some error.
To be exact the cases are : (1,1,22) (2,2,20) (3,3,18) (4,4,16) (5,5,14) (6,6,12) (7,7,10) (8,8,8) (9,9,6) (10,10,4) (11,11,2) 11 cases totally . U subtract (8,8,8) as it's (a,a,a) case so u get 11-1= 10 (a,a,b) cases
@@turkishnaxal1703 3 groups means each group should have atleast one person. lets take an example that there are 6 persons and we have to make 2 groups then both groups are said to be group if atleast one person is include in it.
sir in the 37 indentical books question why you have find ordered to unordred solution for A+B+C= 37 INSTEAD of A+B+C= 28 please give some insights on this
Yes u r right but in previous videos sir has been solving this in another's way ex first if a + b+c = 20 and atleast 3 condition is given we solve for a+ b + c = 14 total solution would be 16c2 which contains ordered solution and we want to remove those ordered solution we only operate on a+b+c =14 only but here sir did for original equation...
In question 2 the no of (a,a,b) including 1 and 2 comes to 18 solutions but sir you have written 17 solutions Please can someone explain me where am i going wrong
Because there are 3 groups, so after subtracting, there are all the unique cases (a ≠ b ≠ c) of 3 groups in different permutation. No of ways to arrange each unique set (a, b, c) is 3!, which is 6. We are dividing by 6 because all 6 are counted in the ordered solutions, and we need them only once. Suppose if it was 4 groups, then we would divide by 4!, after subtracting. In that case, to subtract also there would be three cases ((a,a,a,a) x 1, (a,a,a,b) x 4!/3! and (a,a,b,b) x 4!/(2! * 2!)). Then we can add all these cases, once each.
If we are converting ordered to unordered in last question doesn't that means 3 groups should be identical i.e, having same number of people? Please clarify!
To make it easier, in Ques2, you can simply take a+b+c = 28 (where a, b, c >= 0) hence, we can have cases from 0,0,28 to 14,14,0..making it total 15 unordered cases or 15*3=45 ordered cases. I hope this helps. Cheers!
Similarly, for ques 3, you can take a+b+c = 21 (where a, b, c >= 0) hence, we can have cases from 0,0,21 to 10,10,1...making it total 11 unordered cases, which includes 7,7,7 thus, (11-1)*3=30 ordered cases. I hope this too helps. Cheersagain!
Yeah , i did the same
On @@bhargaviagrawal7216 's Note. If you solve by 2A+B method. You would only get 10.5 on keeping C as 0. So we tend to go to 10. But remember A can have 0 . By that method also we will get 11 unordered cases.
But what about when a =10 , b =10 case it automatically make c =10
But what about when a =10 , b =10 case it automatically make c =10
@ 10:52 instead of taking the equation as A+B+C=37, we can also take a+b+c=28, where a=b
=> 2a+b=28, where a can take 15 values from 0 to 14, thus counting 15 values of the digits repetitive case.
But what about when a =10 , b =10 case it automatically make c =10 ???
@@KjModiwhen a=10,b=10 then c will be c=8 in the above approach not 10. This approach is correct and gives the same answer.
Yes sir.. Please upload remaining videos.. You have claimed to teach 100% cat syllabus.. Please post those videos asap.. Thanks alot.
ABSOLUTE GEM OF CONTENT...
Your way of explanation made the concepts much easier...Thank you sir 😊
Very well-created and systematically explained videos..can't thank you enough !! you made this topic easier to learn :)
Best Teacher ever!!!!!!!!!
Sir, Please upload rest of the lectures of permuations and combination.
Thankyou sir for these videos.
Yours faithfully
YOU ARE AWESOME SIR THANK YOU
19:52 since 2A+C=24 A can take values from 1-11 = 10 values(cannot have 0), and 3A = 24 in this case A=8 we consider it as 1 possiblity but why we are not subracting it from that 10 values since it also have the value 8 in it?
2A + C= 24, A has 11 values from 1 to 11 after subtracting (8,8,8) it becomes 10.
There are 22 videos of p and c.. search permutations and combinations rodha and you will get the complete playlist
Thank you sir!! Very helpful.
you are fantastic Sir
Thank you sir!
thank you Guru ji 🙏
Thank You So much Sir❤
sir at 18:46 , won't a group have minimum 2 people ? thus A,B,C>=2 initial condition.
thanks in advance ^^
I have same doubt
In last question a,b,c can take value=0 , as marbles is to be distributed in 3 groups and marbles can take 0 value
Hi chauhan ,,here we are not distributing the 24marbles to 3 groups ,we are making 24 marbles into 3 groups,so whiling making groups we cant assig 0 ,bec group means atleast 1 ,so hope u understand:)
@@satishchodisetti7896 ah language was confusing. this explains so much better thanks
in question where we distribute 24 identical marble to 3 groups why we need to give atleast 1 to a b c its marble that we are distributing not people so even if group can not have 0 people they can have 0 marbles . if somebody can explain please do.
Yeah having the same doubt😕
@@turkishnaxal1703 if you put 24 marbles on a table and distribute them in 3 groups you will need atleast one marble in a group. if there is no marble, the group does not exist at all ,you will see only two groups on the table.
@@someshjaiswal1410 the groups could be pre existing, they dont have to be decided based on marbles
Sir how all boxes in question 1 are zero there must be one box which will contain balls
Can any one explain me plz
In how many ways can 9 identical balls be distributed among four baskets such that each basket gets at least one ball? what would be the answer for this ?
Are the baskets identical??
Absolutely fantastic approach to provide an edge over other aspirants. I have a query sir , If in the question (similar to similar) we have to distribute among more than 3 boxes, bags , groups etc. Then, how will we approach the problem because then while we convert ordered to unordered cases would be more not just (a,a,b) and (a,a,a). Wouldn't it be a chaos if the question is like how many ways 6 identical books can be distributed in 8 identical boxes.
If we approach with A+B+C+D+E+F+G+H = 6. It won't be possible to convert 13C7 into unordered.
6 identical books would be distributed to 8 identical boxes in 8^(5) ways
As each of the 6 books can be placed in 8 boxes
It is said in p&c 16 video
Sir in the question - No of ways of distributing 37 identical books to 3 identical bags such that each bag contains atleast 2 books, here in (a,b,c) cases there will be some cases where *a* , *b*, *c* will be less then 3 i.e 1, 2, 34 so why you haven't subsracted those cases??
There will be no such cases as we already give 3 books to each bag
One video of introduction of order and unorder is Missing , please upload
In the last question for the condn (a,a,b) shouldnt we need to subtract 1 from 10 because 2(8)+8 is (a,a,a) ?
He did that .....From 1 to 11 there are 11 solutions then he subtracted 1 from 11
20:52 wouldn't there be 11 in place of 10? I am confused.
1 to 11 is = (11-1) + 1 = 11 but as there is 8 which makes the possibility of (a,a,a) and is already included. thus we minus 1 from 11 and have 10
There is some mistake in question... Of 37 books in three bags... The answer( 30C2-3*5)/6 is not integer... How's that possible... There must be some error.
Check your calculation...you might have done a mistake...
Sir combinations give onlycombinations then how is that they are giving ordered pairs
In marbles question....(a,a,b)=11... Pls explain
To be exact the cases are :
(1,1,22)
(2,2,20)
(3,3,18)
(4,4,16)
(5,5,14)
(6,6,12)
(7,7,10)
(8,8,8)
(9,9,6)
(10,10,4)
(11,11,2)
11 cases totally . U subtract (8,8,8) as it's (a,a,a) case so u get 11-1= 10 (a,a,b) cases
17:12 why are we giving 1 marble to each grp?... why does it matter how many people in the grp?
There is no condition to give atlest 1 marble to each grp😕
@@turkishnaxal1703 3 groups means each group should have atleast one person. lets take an example that there are 6 persons and we have to make 2 groups then both groups are said to be group if atleast one person is include in it.
Why A ranges from 0to 9 only?
Sir shouldn't the group have atleast 2people? (16:50)
hi did you confirm it with someone
Nope.....It can be 1 person in 1 grp (like one man army)
Same thinking arose in my mind
In which video was changing from ordered to unordered done? Can anyone tell me the name of the video
please and thank you
in 18
Lecture 9,14,15,17 are missing. Lecture 17 is very-very necessary please provide it sir.
lecture 17 ua-cam.com/video/p5uqB2WUdfQ/v-deo.html
how can only one person be present in a group?
sir in the 37 indentical books question why you have find ordered to unordred solution for A+B+C= 37 INSTEAD of A+B+C= 28 please give some insights on this
coz they have given each box should have atleast 3 books (37-9=28)
Yes u r right but in previous videos sir has been solving this in another's way ex first if a + b+c = 20 and atleast 3 condition is given we solve for a+ b + c = 14 total solution would be 16c2 which contains ordered solution and we want to remove those ordered solution we only operate on a+b+c =14 only but here sir did for original equation...
can anyone tell how to solve question if have to distribute to more than 3 identical things
Sir minimum value of group would be 1 or 2
That'll be 1 only as if u divide 10 people in 2 groups u can keep 9 in one group and 1 in another
In question 2 the no of (a,a,b) including 1 and 2 comes to 18 solutions but sir you have written 17 solutions
Please can someone explain me where am i going wrong
2A+C=37. Each bag must contain atleast 3 books, so C can't be less than 3. Put C=3.
2A+3=37
2A=34
A=17
guys anyone how can we solve different to similar without cases? if sir has explained in some video let me know please
Why are we dividing by 6 all the time???
Because there are 3 groups, so after subtracting, there are all the unique cases (a ≠ b ≠ c) of 3 groups in different permutation. No of ways to arrange each unique set (a, b, c) is 3!, which is 6. We are dividing by 6 because all 6 are counted in the ordered solutions, and we need them only once.
Suppose if it was 4 groups, then we would divide by 4!, after subtracting. In that case, to subtract also there would be three cases ((a,a,a,a) x 1, (a,a,a,b) x 4!/3! and (a,a,b,b) x 4!/(2! * 2!)). Then we can add all these cases, once each.
Sir , I think there should be atleast 2 people to one form 1 group @17:08 plz rectify this issue thank you 🙏🙏
did you confirm it
No 1 enough
how can we say combination to be ordered...whats the logic behindt this. pls someone tell
Because of identical things
Sir unable to understand this video as well. Please upload the video or ordered and unorderd. Thank you so much.
How all identical assumed to all different .... Can someone explain plz
initially, we are assuming it....then we will convert them to unordered form.
group means atleast 2 i guess, but u said atleast 1 how ??????
hi did you confirm it
If we are converting ordered to unordered in last question doesn't that means 3 groups should be identical i.e, having same number of people? Please clarify!
Thank you Sir🙏.