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How do I count the number of selections of size $x$ from a set of $y$ objects where there may be more than $1$ of some object type, where these duplicates are distinguishable but I never have more than $1$ of each.

e.g. If I have a set of $13$ objects:

$A1,B1,C1,D1,E1,F1,G1,H1,H2,I1,I2,I3,I4$

How do I count the number of valid ways to select $6$ objects from this set where I never have more than $1$ object of the same letter, assuming the order doesn't matter and there is no replacement.

e.g.

$A1,B1,C1,D1,E1,F1$ == VALID
$A1,B1,C1,D1,E2,F1$ == VALID
$B1,C1,D1,F1,H2,I4$ == VALID

but

$A1,B1,C1,D1,H2,H3$ == NOT VALID

etc.

Thanks for your input in advance.

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  • $\begingroup$ What is your thinking so far ? $\endgroup$ Feb 2 '18 at 14:43
  • $\begingroup$ Well, I am actually trying to solve a slightly different problem but this is the piece of the problem that I am not sure about. The overall problem was basically a lottery with x winning tickets, from y total tickets where people can hold 1 or more tickets but where any person can only win once (i.e. once one of your tickets wins the others are void). If I know the number of people and how many tickets they have I want to calculate the probability that each person wins. This was a sub-set of my approach to that. $\endgroup$
    – user527452
    Feb 2 '18 at 14:49
  • $\begingroup$ Your edit to your comment gives the matter an entirely different complexion. Are $A, B, C,... H, I$ individuals holding $1,1,1,...2,4$ tickets, or what ? Unless you pose a clear question, you can't get a clear answer. $\endgroup$ Feb 2 '18 at 15:24
  • $\begingroup$ Apologies, I should have just left the question as it was. I will post my overall question separately as I suspect there is a simpler solution. $\endgroup$
    – user527452
    Feb 2 '18 at 16:08
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Make cases, in this instance none from type $H,I,\;$ one from type $H$ only,$\;$ one from type $I$ only, and one each from types$\;H,I$ to get

$\binom76 + \binom21\binom75 + \binom41\binom75 + \binom21\binom41\binom74$

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  • $\begingroup$ Thank you. This answers my question. $\endgroup$
    – user527452
    Feb 2 '18 at 16:07
  • $\begingroup$ You are welcome ! $\endgroup$ Feb 2 '18 at 16:09

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