# Combinations problem - Finding the number

I would like to ask for your confirmation to my thought in regards with the following exercise: One man has 30 different statues, 27 genuine and 3 fake. He sold 10 of these statues in museum A, 10 in museum B and 10 in museum C. What is the probability that each of these museums bought fake statue from the man? I found the probability for each museum, through combinations, that is : $$\frac{{3 \choose 1} {27 \choose 9}+{3 \choose 2 } {27 \choose 8}+{3 \choose 3} {27 \choose 7}}{{30 \choose 10}}$$ Is my way of thinking correct?

Thank you very much in advance.

• I don’t think it is. You are adding up also the cases where one museum recieves more than one fake statue, leaving not enough for the other two. – Tavish Apr 14 '20 at 7:56
• Can you explain your thinking? What are the cases you're counting with the terms in the numerator, and what is the $\binom{30}{10}$ in the denominator counting? – Karl Apr 14 '20 at 8:13
• I think @Tavish is correct. Each museum gets a fake staue means that they all get exactly one fake statue because there are $3$ statues. This means that the probability should be $$\frac{\binom{3}{1}.\binom{27}{9} + \binom{2}{1}.\binom{18}{9} + \binom{1}{1}.\binom{9}{9}}{\binom{30}{10}}$$ – Aditya Apr 14 '20 at 8:16
• @Karl, the $\binom{30}{10}$ is the total number of outcomes.(The probability is asked in the question). – Aditya Apr 14 '20 at 8:18
• Probability that a particular museum purchased fake statue(s) is that, correct. For probability that all museums purchased a fake statue, it is $\frac{3!\times\binom{27}{9\ 9\ 9}}{\binom{30}{10\ 10\ 10}}$ – Rezha Adrian Tanuharja Apr 14 '20 at 8:18

As discussed in comments, your approach is not correct. A correct method would be to assume without loss of generality that the statues are identical except for fakeness and lined up in random order, whereupon the first, middle and last groups of $$10$$ statues are given to the three museums.

The number of distinct ways to select where the fakes lie in the line is $$\binom{30}3$$. The number of ways to select the fakes so that each museum gets one fake is $$10^3$$. The probability is $$\frac{10^3}{\binom{30}3}=\frac{50}{203}$$.

• I really hate this argument with identification of different things. But fortunetly we can avoid it here! – Aqua Apr 14 '20 at 8:54
• @Aqua I prefer doing this because it tends to give small numbers, which I can more easily run through my head. – Parcly Taxel Apr 14 '20 at 8:56
• @Aqua In other words, here I am assuming that the man has $30$ Polteageists and $3$ of them are Phony Form. – Parcly Taxel Apr 14 '20 at 8:57

Your answer seems wrong. Although, from your answer I am unable to judge about how you exactly arrived at it but here's a hint to how you could have approached the problem:

Just in case you don't know: Number of ways to divide n objects in n1 groups of m1 object, n2 groups of m2 object and so on till nk groups of mk objects such that Σni mi = n can be given as: $$\frac{n!}{(m1 !)^{n1} (n1 !) .... (mk !)^{nk} (nk !)}$$

(This result can be arrived at using simple product rules and a bit of intuition or more formally by using set theory and providing appropriate bijections)

You can use this to frame your answer with the following thought process:

Compute total possible ways to distribute( without any restrictions ) and use this as denominator of your probability fraction.

Compute total possible ways to distribute 27 genuine pieces among 3 groups of 9 each. Also, calculate total possible ways to distribute 3 defected pieces among 3 groups of 1 each. And multiply these two computations using product rule to get total number of ways to distribute 30 pieces such that each one gets 9 genuine and 1 defected piece. This gives you the numerator.

Finally, obtain the probability in terms of a fraction.

Hope this helps!

• Thank you very much! – kostas Apr 14 '20 at 9:20
• @kostas consider upvoting the solution if it was helpful – SAGALPREET SINGH Apr 14 '20 at 12:49

Another answer (related to Parcly's): Imagine each museum has 10 tables, one for each statue it bought. The first fake occupies one of these tables. The next fake is equally likely to occupy each of the remaining 29 tables, so the probability that it's in a different museum from the first fake is $$\frac{20}{29}$$. In this event, of the 28 remaining tables, 10 are in the last museum, so the probability that the last fake is in a different museum from both of the first two fakes (given that the first two are in different museums) is $$\frac{10}{28}$$. The probability that this all happens is therefore $$\frac{20}{29}\frac{10}{28}=\frac{50}{203}$$.

• Thank you very much!! – kostas Apr 14 '20 at 20:48