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Consider the following theorem:

Let $A$ be a set of size $n$. Then there are $\frac{n!}{k_{1}!\cdot\ldots\cdot k_{t}!}$ ways to partition this set into $t$ nonempty set of sizes $k_{i}$.

And consider the following example:

There are $36$ cards, with half of them being blue and half being black. These cards are shuffled and then divided into two equal stacks of $18$ cards. One has to determine the probability that in both stacks there is the same number of blue and black cards.

My question is: How can this example by solved with the above theorem ?

The answer is $\frac{\binom{18}{9}\binom{18}{9}}{\binom{36}{18}}$ - but I got to it with my own solution which doesn't use the above theorem, which bothers me, since in the notes I'm reading, this example comes as a direct application of the above theorem.


For anyone who cares, heres my solution: If the number of blue and black cards has to be identical in both stacks, we have to have $9$ blue and $9$ black cards in both decks. Since there are $18$ blue and $18$ black cards totally, there are $\binom{18}{9}\binom{18}{9}$ possibilities for one stack. But if we select one stack, the other is automatically determined, so there are actually $\binom{18}{9}\binom{18}{9}$ possibilities to select both stacks. Since totally there are $\binom{36}{18}$ ways to divide the cards into arbitrary stacks of the same size, we get the answer).

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Your solution is better. And the formula you were given counts ordered partitions, not partitions. –  André Nicolas Dec 26 '12 at 17:36

1 Answer 1

up vote 1 down vote accepted

For the numerator, from the theorem there are $\frac{18!}{9!\cdot 9!}$ ways to partition the $18$ blue cards into two (ordered, as André Nicolas commented) sets of $9$ blue cards and $9$ blue cards, and similarly for the black cards.

For the denominator, from the theorem there are $\frac{36!}{18!\cdot 18!}$ ways to partition the $36$ cards into two sets of $18$ cards and $18$ cards.

Hence the result.

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