6.1. Three pirates, Joe, Bill, and Tom, found a treasure containing 70 identical gold coins, and they want to divide them so that each of them gets at least 10 coins. How many ways are there to do this?
Solution. Let the treasure consist of $n=70$ coins and each pirate should receive no less than $k=10$ coins.

Give each pirate $k-1$ coins, and lay out the remaining $n-3 k+3$ coins in a row. To divide the remaining coins among the pirates, it is sufficient to place two dividers in the $n-3 k+2$ spaces between the coins. Thus, Joe will receive the coins to the left of the first divider, Bill the coins between the two dividers, and Tom the coins to the right of the second divider. The number of ways to place these two dividers is $C_{n-3 k+2}^{2}=\frac{(n-3 k+2)(n-3 k+1)}{2}$.

If $n=70, k=10$, then we get $C_{42}^{2}=861$ ways.

Answer: 861.