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Continuing with self-study probability,


There are 16 army troops that need to protect 8 oil towers (numbered from A to H). They lost contact with each other. Supposing that each army troop has equal probability to protect any of the tower.

Event B: Which is the probability of exactly 2 of the towers to remain without army troops protecting them.

Attempted Solution:

I try to solve the problem using these recently acquired heuristic and my attempted solution is this one:

$ \binom{2}{0} * \frac{\binom{16}{6}}{\binom{16}{8}} = 0,62$

Meaning taking 0 of 2 towers and getting the 16 troops into the 6 remaining towers, but I'm pretty sure I'm wrong (prob = 0.62 which seems a little higher than expected).

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Hi, those problems seem interesting, can you give me a source please. Thanks – Sebastian Griotberg Aug 22 '13 at 7:53

There are $8^{16}$ possible arrangements. Now we must figure out how many of these are such that exactly two are empty. Suppose A and B are empty. Then there are $6^{16}$ ways to distribute the rest of the soldiers, but some of these have other empty oil towers. We use the inclusion-exclusion principle: $\displaystyle\Big[6^{16}- \binom{6}{1}5^{16}+\binom{6}{2}4^{16}-\bigg(\binom{3}{2}-1\bigg)\binom{6}{3}3^{16}+\bigg(\binom{4}{3}-1\bigg)\binom{6}{4}2^{16}-\bigg(\binom{5}{4}-1\bigg)\binom{6}{5}1^{16}\Big].$

To get the final answer, you have to multiply this by the number of ways of choosing $2$ from $8$, and then divide by the total number of options:

$\displaystyle\frac{\binom{8}{2}}{8^{16}}\Big[6^{16}- \binom{6}{1}5^{16}+\binom{6}{2}4^{16}-2\binom{6}{3}3^{16}+3\binom{6}{4}2^{16}-4\binom{6}{5}1^{16}\Big]\approx \frac{1}{5}.$

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