Re: Groupings and Distributions  
hi tg.this is a wonderful article.this was a weaker area of mine,but your article has helped clarify the concept.three cheers 2 u. 
Re: Groupings and Distributions  
Excellent!!! I wish we could stop time so that we'd have ALL your wonderful articles before CAT 07! 
Re: Groupings and Distributions  
your relentless marvellous work will help many cat aspirants in their preparations......................thanx dude.keep up the good work !! 
Re: Groupings and Distributions  
Hi TG Please explain how 1 1 and 3 groupings can be distributed in 3! ways as 2 groupings are similar.This should be 3!/2! ways.same for 1 2 2. 
Re: Groupings and Distributions  
Hi TG, Brilliant boss! This is my first visit to the site and already i think its gonna be very useful! Hats off to the your illustrations. Cheers

Re: Groupings and Distributions  
Ur intelligence and ur creativity is makin me jealous...the way u xplain and the way u make difficult thins so simpler..thas really awesome.. 
Re: Groupings and Distributions  
Gr8 article... the best thing abt TG is that he explains complex things in a simple manner which even the coaching instis are unable to do. Hats off TG!! 
Re: Groupings and Distributions  
Really good article. You are doing a great job and please try and get in as many such articles as possible before CAT 07. 
Re: Groupings and Distributions  
Just awesom!

Re: Groupings and Distributions  
Hi Fics, is the case (a, b, cde) same as (b, a, cde)? Total Gadha 
Re: Groupings and Distributions  
Thankyou sir Point noted 
Re: Groupings and Distributions  
sir can u please clarify on this point 
Re: Groupings and Distributions  
Fabulous Article TG, Thanking you............... I bow to TG 
Re: Groupings and Distributions  
Please TG Sir include a chapter on Permutation&Combination 
Re: Groupings and Distributions  
Hi Shreshtha, At least what I have written. I have covered this case in "boxes different balls same". Total Gadha 
Re: Groupings and Distributions  
HI TG In the example that you have explained 5 balls and 3 boxes finding the groupings of the ball were easy since the number of balls are less,where as when the number of ball are more say for example 40 balls and 10 boxes its becomes tedious to find the groupings of the ball.Your explanation have cleared the concept but is there any shortcut for solving the 1 and III condition as we have formula for the II and the IV condition. I the boxes are similar and the balls are similar. Eagerly awating for your reply 
Re: Groupings and Distributions  
Hi TG, Can you pls throw more light on the approach to this particular type of distribution problem. There is a guy named Gadhanand...who appeared an exam that consisted of 4 papers...the maximum marks of 3 of these papers is 50...while the max of 4th paper is 100... In how many ways can he secure 60% of marks in the exam in order to change his name from Gadhanand to Vidyanand??? Partha 
Re: Groupings and Distributions  
HI TG , Thanks a lot , you have covered each and every nuances of the particular problem and that is what keeps me in awe 
Re: Groupings and Distributions  
Hi TG, i am quite a student of yours and i like your teaching methods. i have a problem here that has been puzzling me for quite some time now. here is how the problem goes, in how many ways can 12 similar chocolates be distributed among 3 persons where each person can get a maximum of 6 chocolates? a similar problem is as follows, find the number of four digit numbers that can be generated using the digits 0,1,2.... 9 such that the sum of the digits of the four digit number is 12. I feel the solution to these two should be on the same lines . please let me know, i also welcome other students to gimme the solution if they know. regards, a student 
Re: Groupings and Distributions  
Hi Nice Smile, In how many ways can 12 similar chocolates be distributed among 3 persons where each person can get a maximum of 6 chocolates? You can solve this question by using the reverse method for distribution that is picking. First distribute 6 chocolates each to all the three students. Now they have 18 chocolates in total. Now you need to pick 6 chocolates out of these 3 groups which is the solution of a + b + c = 6 = 8!/6!2! = 28 Find the number of four digit numbers that can be generated using the digits 0,1,2.... 9 such that the sum of the digits of the four digit number is 12. As the first digit cannot be 0, we require the coefficient of x^{12} in the product (x + x^{2} + x^{3} + ... + x^{9})(1 + x + x^{2} + x^{3} + ... + x^{9})^{3} = coefficient of x^{11} in the product = (1  x^{9})/(1  x) × (1  x^{10})^{3}/(1  x)^{3} = (1  x^{9})(1  x^{30}  3x^{10} + 3x^{20})(1  x)^{4} = 332 (expanding (1  x)^{4}). I hope the answer is correct. Total Gadha 
Re: Groupings and Distributions  
Hi TG, I liked the solution to the first question. Thank u so much, i never thought about reverse distribution. now i m enlightened with this too i did not understand ur solution to this question: Find the number of four digit numbers that can be generated using the digits 0,1,2.... 9 such that the sum of the digits of the four digit number is 12. wy do we take the coefficient of x^11 as the answer. i do not understand wy are u multiplying those two expressions in the first place . please explain the logic behind it. but i did try another way to find the answer and it seems correct to me. please verify. the sum of the numbers should be 12 and i consider them as 12 balls. and since it is a 4 digit number i consider it as four different baskets(a,b,c,d). Now, the first digit cannot be zero so i put one ball in the first basket(a){let me call the rest of the balls that go to basket one as a'}. therefore the problem reduces to the number of ways of dividing these 11 balls into the four baskets with three important caveats: 1. all the cases where all the 11 balls have been put in the same basket have to be excluded 2. all the cases where 10 balls have been put in the same basket have to be excluded 3. all the cases where 9 balls have been put in the first basket has to be excluded(since it ll make the value as 10 for the first basket a, this is invalid since the maximum that we can put in a basket is 9). Now for the calculations, no. of ways of dividing 11 balls among 4 baskets is: a'+b+c+d = 11, (n+r1)C (r1)= (11+41)C (41) = 14 C 3 = 364 Now for case 1: if all eleven are put into one basket it is wrong since the maximum that can be there in a basket is 9, therefore we exclude this. this is nothing but arranging 11,0,0,0 which is equal to 4!/3! = 4 ways similarly for case 2: this is nothing but arranging 10,1,0,0 which is equal to 4!/2! = 6 ways for case 3: we need to consider only those cases where we have 9 in the first basket, and they are the number of ways of arranging 9,1,1,0 (with nine fixed in the first place) = 3!/2! = 3 ways 9,2,0,0 (with nine fixed in the first place) = 3!/2! = 3 ways hence the total is 364  (4+6+3+3) = 36416 = 348 the answer is different from what you have got, but i feel this should be right. Please confirm. 
Re: Groupings and Distributions  
Hi Nice Smile, Both of us made a mistake, the answer is 342. You have calculates 4!/2! = 6 whereas it is equal to 12. Total Gadha 
Re: Groupings and Distributions  
thank you tg, for informing me that my approach is right. i have been waiting the whole day for your reply but i still did not get how you solved it ..... and i missed clearing a cut off once in cat just because of an addition mistake (a silly mistake like this one) ... i have to wait an year more coz of that ..... can you please tell me why did you multiply both of those expressions?? 
Re: Groupings and Distributions  
Hi Nice Smile, Simply because of the fundamental rule of counting that if a job can be done in m ways and the other in n ways, both can be done in m × n ways. The expression (x + x^{2} + x^{3} + ... + x^{9}) denotes the values 1st digit can take. The expression (1 + x^{2} + x^{3} + ... + x^{9}) denotes the values 2nd, 3rd and 4th digit can take. Total Gadha 
Re: Groupings and Distributions  
thank u tg.. i got it now 
Re: Groupings and Distributions  
Hi TG Still not getting that expression funda. Can you elaborate bit more. 
Re: Groupings and Distributions  
hi kamal, the logic goes like this .... whenever we are considering the power of x, then it is like chosing one among the digits 09, therefore , x^0 stands for 0, x^1 stands for 1 and so on. to get the sum of the digits as 12 we check the coefficient of x^12 which means that the sums of the terms was 12. This works since the coefficient of x^12 will have all the combinations of x^n such that it gives the answer. for ex: x^12 can be written as x * (x^9) * (x^2) * (x^0) and this would represent number 1920, where x is from first expression, (x^9) is from second expression and so on. we did not take (x^0) for the first expression since the first digit cannot be zero. hope this clarifies ur doubt. 
Re: Groupings and Distributions  
In how many ways can 10 identical presents be distributed among 6 children so that each child gets at least one present.....pls solve 
Re: Groupings and Distributions  
hello TG sir..i am new to your site and am interested in buying the geometry ebook..what are the topics covered in it sir?is it a comprehensive book for cat?similar doubts for the numbers e book as well..and regarding the cat cbt club will it include full length mock tests as well? 
Re: Groupings and Distributions  
hi tg, can you kindly explain the following case in detail i'm not able to grasp from the given explanation IV. the boxes are different and the balls are different. thanks 
Re: Groupings and Distributions  
ignore my prev post ....got the logic..... 
Re: Groupings and Distributions  
Find
the number of four digit numbers that can be generated using the digits
0,1,2.... 9 such that the sum of the digits of the four digit number is
12.As the first digit cannot be 0, we require the coefficient of x^{12} in the product (x + x^{2} + x^{3} + ... + x^{9})(1 + x + x^{2} + x^{3} + ... + x^{9})^{3} = coefficient of x^{11} in the product = (1  x^{9})/(1  x) × (1  x^{10})^{3}/(1  x)^{3} = (1  x^{9})(1  x^{30}  3x^{10} + 3x^{20})(1  x)^{4} = 342 (expanding (1  x)^{4}). please explain how did you calculate the value 342 .... i really did not understand 
Re: Groupings and Distributions  
awesomeness redefined.......... kindly post more articles, m not seeing any new article now plz dont hibernate, we need u in this monsoon 
Re: Groupings and Distributions  
Hi TG, In response to your answer to "In how many ways can 12 similar chocolates be distributed among 3 persons where each person can get a maximum of 6 chocolates?" It's similar to when balls are similar but boxes are different. That way number of distibutions should come a multiple of three( because we are distributing to three people). But 28 is not multiple of three? 
Re: Groupings and Distributions  
thanks TG sir.really helped.always searched permutation and combination.but now got a clear method of solving these questions on basis of grouping and distribution. 
Re: Groupings and Distributions  
Hi, It's a nice example on group and distribution. However I have one question on this topic. In case II : i.e. if the boxes are different but balls are simillar, the formula for finding total no of ways of distribution of n nondistinct objects into r distinct groups is ^{n+r1 }C_{r1.} In case IV : i.e. if the boxes are different and balls are different, the formula for finding total no of ways of distribution of n distinct objects into r distinct groups is r^{n}. Is there any straight forward formula for finding out total no of ways of distribution for Case I and Case III (Like Case II and Case IV) ?

Re: Groupings and Distributions  
how can we group 9 members in 3 groups of three 
Re: Groupings and Distributions  
9!/(3!)^4 
Re: Groupings and Distributions  
its not perfectly divisible and the no of ways can not be in fraction!!!! 
Re: Groupings and Distributions  
Hi Nitish 9!/(3!)^{4} = 280 i.e. an integer. Please check. Kamal Lohia 
Re: Groupings and Distributions  
Dear Sir, I am not able to get this concept which you have also asked for Could you please explain the same? 
Re: Groupings and Distributions  
Hi Sir, Please let me know is "Case IV: the boxes are different and the balls are different " similar to "grouping n different things among r groups =r^n" ?
Thanks.

Re: Groupings and Distributions  
In how many ways can 12 similar chocolates be distributed among 3 persons where each person can get a maximum of 6 chocolates? TG Sir, suppose if this question was changed such that, the first person receives at most 6, second person receives at most 7 and third person receives at most 8. Then this would be about solving A+B+C = 12 where A is less than or equal to 6 B is less than or equal to 7 C is less than or equal to 8 Now, what about this case sir? If the technique mentioned by you is used, then we have to distribute 6 to A, 7 to B, 8 to C. So in total they have 6+7+8 = 21 chocolates. So we have to pick 9. So this is about solving the equation A+B+C = 9. This is by the method of partition gives 11C2 ways or 55 ways. But the solution mentioned says 45 ways What mistake am I doing ? 
Re: Groupings and Distributions  
Hi Gautam, I already answered the same problem. 
Re: Groupings and Distributions  
Hi TG Sir and all, This post clearly explains the following 1. Similar Things > Different places 2. Similar Things > Similar places 3. Different Things > Different places 4. Different Things > Similar places There is also a great explanation about 1. Grouping of Different Objects a. into groups of equal size b. into groups of UNequal size 2. Grouping of Similar Objects But, Can some one pls explain how to GROUP AND DISTRIBUTE cases where both SIMILAR AND DIFFERENT objects are present. Eg, the case where Balls > B1, B1, B1, B2, B3 have to be grouped and distributed ? Anybody .. pls help me out ! 
Re: Groupings and Distributions  
Hi Ankit sir, Thanks so much for the reply!! ... I've been looking for this explanation all over the place I have a few doubts from what you said.. for the first question which TG had explained in the forum 1. In the question which TG had solved, A+B+C = 12 where A,B,C can take a maximum value of 6. This means, that A,B,C are all less than or equal to 6 right ? 2. Suppose if I changed the limits to  > A,B,C are all less than or equal to 5. Even then the technique doesn't give the correct answer. The equations to be solved will be A+B+C=3 I assumed that this technique (mentioned by TG) could be used only if all 3 variables had the same upper limit. Am i wrong in thinking so ? (because in this we are removing a single 3 from each of A,B,C. The upper limits are 5. SO we can definitely remove a 3.. But it doesn't work) 3. Can you please explain how you arrive at the variables X,Y,Z ? 4. Can you throw more light on your statement  > "But in this case method we can not take 9 or 8 or 7 back from x, same for y or z" Thanks so much for the explanation. Pls help me in clearing this up 
Re: Groupings and Distributions  
awesome article sir Please suggest me some more articles on Permutations and combinations 
Re: Groupings and Distributions  
boxes are different...let 0,1,4 be placed in respective boxes...say A,B,C....they can also be put as 0,4,1....1,0,4....1,4,0....4,1,0....4,0,1....0,1,4....like this in A,B,C in 6 ways..i.e. 3!.... 
Re: Groupings and Distributions  
In how many ways can you divide five similar objects (say a, a, a, a, a) into three different groups? When we make groups of similar objects, the only way we can differentiate two different ways of grouping is by differentiating between groups when they have different size (number of objects). Therefore, in case of similar objects, the number of different ways we can group them is the number of different sizes of groups that we can make. Letâ€™s see how many groupings of different sizes we can make for 5 similar objects.
sir i have understood till now.But now can u please explain this concept with few more examples .and doesnt this have any formula as like for dissimilar objects . while coming to distributions i have got the idea that first we need to divide them into different groups and then we can follow your procedure. But how do we find those groups ...?? 0 0 5,0 1 4,.... how did u get this sir ??? hope you reply soon 
Re: Groupings and Distributions  
Hi Sravani You are to distribute 5 identical things into 3 groups. Just think of what groupings are possible and write them. See if you get anything other than those mentioned in above post. To start small, you can just check the ways to distribute 2 identical things (say a, a) in 2 groups. Isn't it, there are 2 ways only i.e. one a in each group i.e. 1, 1 OR both a's in one group and other on empty i.e. 0, 2. Regarding second query, 'formulae' are also application of some external brain. So why to depend on someone else's brain if you have a one and can apply it. By the way, there is not any direct formula for this so far. Kamal Lohia 