Permutation and combination

[kks3]

There are 51 machines in a factory, 204 workers work in the factory.

Each day, grouping of workers is done. A group of 4 workers work on a single machine in a day.

On the second day, the grouping is done again o, the condition that no 4 members who are together on the first day will be together on the second day. On the third day, those who are together on the first and second day should not be together.

The question is:
FOR HOW MANY DAYS IS THIS POSSIBLE ??

[imatfaal]

A quick guess would be that it is workers Choose machine - ie 204 choose 51 see here

 

This is the number of combinations of a set of 51 things chosen from a set of 204, with no repetitions and where order does not matter

 

if we define the full set to be chosen from as n and the number of choices as r and ! represents the factorial

 

[latex]\binom{204}{51}=\frac{n!}{r!(n-r)!}[/latex]

[John]

I'm assuming, given the way you've worded this, that any two or three people who were together on one day can still be together on the next, just not all four. For instance, if John, Jacob, Jingleheimer, and Schmidt were a group the first day, then the second day we could still have something like John, Jacob, Jingleheimer, and Gertrude. In that case, my first guess would be [math]\frac{{204\choose4}}{51}[/math].

My reasoning is as follows: On the first day, we have [math]204\choose4[/math] ways to choose four people given that order doesn't matter. We select 51 such groups to man our 51 machines. The second day, then, we have the original [math]204\choose4[/math] possible groups, minus the 51 we chose the day before. On the third day, we lose the 51 groups we had on day two. And so on. Thus, the number of days we can do this is the number of times we can subtract 51 from [math]204\choose4[/math], i.e. [math]\frac{{204\choose4}}{51}[/math].

If no two people can share a group twice, then the options are, of course, much more limited. On the first day, we still have [math]204\choose4[/math] possible groups. On the second day, we lose not just the 51 groups from the first day, but all the groups containing any two people who were grouped together the first day. This is a bit more complicated to work out after the second day, unless I'm missing something simple.

Edited December 30, 2013 by John

[imatfaal]

Unfortunately I cannot give myself a neg rep!

[studiot]

The title of this thread is permutations and combinations.

 

John and imatfaal have correctly identified that combinations are the appropriate operation here, since order does not matter in this question.

 

John has further identified an ambiguity with the posing of the data.

 

You may wish to know that

 

The number of combinations of n things r at a time in which p things always happen is

 

_n-pC_r-p

 

The number of combinations of n things r at a time in which p things never happen is

 

_n-pC_r

[rktpro]

The number of days such a system can work is same as the number of distinct groups containing 4 members each which can be made out of 204 workers. 204!/[(4!^51)51!]

[rktpro]

I made a little mistake in language. The number calculated would give no of distinct sets of 51 groups containing 4 members each.