Three coins are randomly placed into different positions on a $4×10$ grid. The probability that no two coins are in the same row or column can be expressed as $a/b$ where $a$ and $b$ are coprime positive integers. What is the sum of $a+b$ ?

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Somebody decided to turn to MSE to get their homework done and is rapidly posting it on the site by parts. –  Did Apr 2 '13 at 9:39

What is your probability space (i.e. how many different combinations of the three coins are there)? How many options do you have for the first coin? How many for the second, given that the first one is placed? After the two coins are placed, how many options are left for the third coin?

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Once we place the first coin, there are $39$ available positions for the second coin. But we must avoid the positions that are in the same row or column as the first coin. So imagine rubbing out the row and column of the first coin. That leaves a $3$ by $9$ grid of "good" positions for the second coin.
So the probability that after two placements, we are OK, is $\dfrac{27}{39}$.
Given that we are OK so far, place the third coin. There are $38$ places it could be put. To count the places that avoid the rows/columns of the first two coins, imagine rubbing out these rows and columns. That leaves a $2$ by $8$ grid, so $16% places. So given that the second placement was OK, the probability that the third is OK is$\dfrac{16}{38}$. So our probability is$\dfrac{27}{39}\cdot \dfrac{16}{38}$. Now we are finished, but are forced to do some mechanical work so we can be mechanically graded. Another way: We can choose the locations of the coins in$\dbinom{40}{3}$equally likely ways. Now we want to count the number of "good" choices, in which no row or column has$2$or more coins. The rows an be chosen in$\binom{4}{3}$ways, and for each choice the columns can be chosen in$\dbinom{10}{3}\$ ways. So our probability is $$\dfrac{\binom{4}{3}\binom{10}{3}}{\binom{40}{3}} .$$ Finally, evaluate to get the answer in the form required.