22 thoughts on “Distinct Objects in Identical Groups

  1. I am getting 46 as the answer.1st (5,0,0) in 1 way.then (4,1,0) in 5!/4!=5 ways.next (3,2,0) in 5!/3!*2!=10 and(2,2,1) in 5!/2!*2!=30ways.so total 46 ways.please let us know if I am doing it correctly.

  2. Sharmi – You have made two small mistakes.

    Firstly, 2, 2, 1 needs to be again divided by 2 (see the cases above marked with a comment in brown!) since the groups with 2 and 2 will be identical. So it will be 15 and not 30 ways for that distribution.

    Secondly, there will be a case of 3, 1, 1, which you have not taken into account. This will yield 10 more cases. Taking these two corrections into account, the result will be 41.

    regards
    J

  3. @catcracker – In one of your earlier posts u had mentioned… Number of ways of distributing n distinct objects into r identical groups… is given by (r^n) / r!
    though i got the answer 41 in above case… the answer using the above formula come to 40.5
    plz guide me…. on this topic

  4. it is a very important question… in almost every book this formula is given for dividing distinct objects in identical groups…. but no book has mentioned any examples….

  5. Hi Girish,

    Actually, I had not mentioned this, as I have only talked of distinct groups till last week. The formula you mentioned would work, but one would have to take the greatest integer function for that if I recall correctly – [243/6] = 41 or in the earlier problem [2187/6] = 365. There is some correction required for multiple empty groups, I think. It is somewhat prone to confusion and hence risky to apply (as you found out!), so I am going by the slow and steady method of enumeration. I’ll check if I can find any further clarification on that formula and will comment if I do.

    I generally prefer going by the underlying logic in most cases – CAT rarely asks a direct formula-based question and to solve the twists one has to understand the basic principles. Here, for example, CAT might ask “In how many ways can we distribute 7 distinct objects into 3 identical groups such that no two groups contain the same number of elements?”

    regards
    J

  6. Yes , definitely sir.. this method of listing is far better…
    I have one more doubt sir… the mutual fund problem u had presented before us… which u solved graphically…can u please explain the method in detail…

  7. I assume you are referring to this? https://crackthecat.wordpress.com/2013/01/24/weighted-averages-cat-scan/

    The idea was of a minimax situation. Let me give you a discrete example. Suppose I say, you have 3 choices. If you pick A, you will either win $10 or $100 (equal probability of winning or losing). If you pick B you will either win $40 or $50. If you pick C you will win either win $ 20 or $80. Now the average (expected) earning from A and C are much higher than that from B. If you were to pick on that criterion alone, you would pick A.

    However, if you are risk-averse and wish to maximise your GUARANTEED return, then you will pick B as you are guaranteed at least 40 which is higher than the other “worst-case” scenarios.

    In the given problem, Shabnam can consider investing partly in A and partly in B – if she does, the red line shows her worst-case returns. As you see from the graph, the maximum case is at a point which is higher than the returns she would get from investing in C (which is a fixed value) and therefore to maximise her GUARANTEED returns she should choose that mix of investment.

    There were other ways to do the problem, the graphical approach was just a different way of looking at it 🙂 It becomes like a co-ordinate geometry problem in this approach.

    regards
    J

  8. sir i wanted to ask that if in this case of ” distinct objects in identical groupos” we get a number jahan possibilities ko enumerate krna posible nhi hai unlike in this case jahan humne cases likhw nd fir evaluate kia.. den what to do??

    • Aashika, you would not get a problem like that in a CAT. In fact even this one here is slightly beyond what one could expect 🙂 So don’t worry. They will give something jo manually ho sakta hai, not where you would need esoteric knowledge….

      regards
      J

  9. I don’t understand the question. 7 different people to be divided into 3 identical groups? How can the groups be identical? Groups are named identical if the group size is same and some other differentiating factor (like 3 different sections of a school, 3 hobby classes) is not present.

    • Sanchi, consider one situation where we have 7 people PQRSTUV to be divided into 3 groups A B C. Then PQR in A, ST in B and UV in C is different from ST in A, PQR in B, and UV in C. But if we want to send these people off for a picnic and 3 cars are available, then it doesn’t matter who goes in which car, it only matters who are together. So (PQR), (ST) and (UV) will be only one case instead of 6 cases in the previous example.

      regards
      J

  10. Hi J,
    While grouping 7 people, say in groups of (4,2,1), are we selecting 4 from 7, then 2 from remaining 3, that effectively give us 7C4 * 3C2. And in case we have identical groups, say (3,3,1) we are selecting 3 from 7, then 3 from 4, that effectively gives us 7C3*4C3/2! (division because we have 2 identical groups).. and so on for each case
    Now if we have another situation where we have to divide 6 guys in a groups of (2,2,2) will the case be 6C2*4C2/3! (because we have 4 identical groups)?? am i extending the logic correctly?

    • Yes. You could also do the first part 6C2 * 4C2 directly as 6! / (2!2!2!) (see the post on distinct objects in distinct groups) so the answer here would be 6! / (2!2!2!3!)

      So if you want to divide let’s say 30 people into 6 groups of 5, then it could be done in 30! / [(5!)^6 * 6! ] ways

      regards
      J

  11. In the given solved example (6,1,0) and (6,0,1) should be two distinct cases right? I am asking this since you’ve considered only (6,1,0) as a case and not the other.

  12. What about a case like distinct objects into distinct groups, for example divide ten distinct balls into three different boxes?

  13. What about a case like distinct objects into distinct groups, for example put ten distinct balls into three different boxes, considering the order in which the balls are put in the boxes.

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