Distribution of some or all objects, distinct groups

25 thoughts on “Distribution of some or all objects, distinct groups”

1. himmatwala |

in the last one , 12 fish must be distributed among 5 people , i think there is an error mr. J

• Himmatwala, it is not 12 fish but “not more than 12 fish”. So effectively, as in the previous example, it is like distributing 12 fish among 6 people (one dummy). Hence 6 groups.

regards
J

2. appy |

In the first example shouldn’t it be 26C21 as 21 people are der

• 21 people i.e. groups => 20 partitions….re-read the first post on Theory of Partitioning, it will become clear ðŸ™‚

regards
J

3. Jayavignesh Ravichandran |

I’m not getting that addition of dummy concept…Its confusing.However, the result we are getting it correctly.Formula applied after adding dummy is also clear.But Dummy concept is making me go nuts .Expecting a better explanation sir

• Jayavignesh Ravichandran |

• What’s there to explain? If we don’t have to distribute all, then the ones left over (i.e. which are not distributed) will form one extra group, we are just adding that group.

regards
J

4. Abhijeet |

Hi J,
I have a conceptual doubt i the first problem. When I read the problem my initial impulse was to select “student for coin”. Therefore, first coin can go in 21 ways(one dummy for the case when we don’t have any student more than 60%). Thus the answer should be 21power 6.
I your solution. I know I am confused somewhere between identical and distinct objects. But I can’t figure out exactly which to use when?
Any help will be appreciated.

• If they are distinct then your logic would work (for example, if instead of coins they had awarded medals with the subject name written on it). But if they are identical, then the first coin going to you and the second to me is the same as the first going to me and the second to you. So there will be many such instances of double counting (and indeed multiple counting). So that method won’t work. Having established that, we look for an alternative to use when identical objects are involved, which is provided by the method of partitioning.

regards
J

• Sandeep |

The entire series of posts on PnC have been absolutely amazing. You have a natural gift of imparting knowledge and we are all glad your passion lies in teaching. Thank you!!!

5. Saurabh |

In the questions, as you have introduced a dummy person, then the case becomes that atleast one person will always receive some coins/fish. In that case doesn’t it become a mixture of the two cases explained earlier where we consider a case where it is possible for a group to have a value of zero and the next case in which that is not possible?

• Yes, but that will lead to not just two but many cases. As you can see we had to list 7 cases to solve the first question by that approach.

regards
J

6. Saurabh |

Ok! Thanks. ðŸ™‚

Hi J,
I am following your blogs since the past few days, I must say they are interesting and insightful :). The alternative approaches for a question are really helpful. Keep Blogging. Getting back to the above blog,
from the 1st question, can we induce that,

rCr +(r+1)Cr + (r+2)Cr…. nCr = (n+1)C(r+1)

I tried it for simple numbers and it worked, so just wanted your mark of confirmation on that.

8. Hi J,
I am following your blogs since the past few days, I must say they are interesting and insightful. The alternative approaches for a question are really helpful. Keep blogging. Getting back to the blog above,
from the 1st question can we induce that,

rCr + (r+1)Cr + (r+2)Cr +â€¦ nCr = (n+1)C(r+1)?

I tried it for simple numbers and it worked, so just wanted your mark of confirmation before adding it to my tips and tricks list.

• Hi R,

Yes, that would be a logical deduction, and can be proved by slightly changing the way of writing it.

Think of it instead as rC0 + r+1C1 + r+2C2…and so on.

Then if we remember that nCr + nCr+1 = n+1Cr+1 and that rC0 is the same as r+1C0 (which is 1) then we can prove the result you stated.

So yes, it will always be true. Well reasoned!

regards
J

9. seema |

In first problem, When we are considering a+b+c+d… (20 students) = 6 then aren’t we ignoring the case of more than one coin achieved by same student?
Also, In question it is mentioned that, there is a possibility that no one gets a coin (if no on scores at-least 60%), so by considering a dummy, someone will definitely get one. So that case is ignored..
Also taking a dummy ignores the case of achieving more than one coin by same person..? Isn’t it? I might be getting it wrong..plz explain

We could have a = 6, b = c = d = … = 0. So even all the coins could go to one person.
If we consider the dummy, then there is also the possibility that the dummy gets all 6 (which means none of the original 20 people get any, all the coins remain with the teacher who was awarding the coins)
And even with the dummy we could have a case like a = 4, b = 2, c = d = … = dummy = 0. Or for that matter a = 6 and all others including dummy = 0. So one person can be getting multiple coins here too.

regards
J

11. Hey, could you please explain if we are treating words like- up to and at least, in the same way, i,e adding a dummy and then partitioning?

• No because the letters are distinct there. This is for identical objects.

regards
J

• Sorry, but what is distinct? Partitioning is used for identical objects only, right?

12. Which is precisely my point. A, B, C, D etc are distinct right? So I cannot use partitioning but have to use the equivalent of n^r or nPr logic (depending on whether repetition is allowed or not) as demonstrated in these earlier posts:

regards
J

• I think I wrongly phrased my query. I meant that questions which have conditions where up to and at least is used, we use dummy variables in such questions, right? Ex- UPTO 20 elephants, or, AT LEAST 60% marks.
Sorry for the misunderstanding!

13. Yes, thank you so much ðŸ™‚