There are 70 beads.~ $$A$$ and $$B$$ take turns taking beads from the pile ($$A$$ goes first). Each person can take between 1 and 4 beads per turn. The person who takes the last bead loses. If $$A$$ wants to win, how many beads should he take on the first turn? 
Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
$$(70-1)\div (1+4)=13\cdots 4$$  $$A$$ takes four bead first, and then regardless of how many beads $$B$$ takes, as long as the sum of $$A$$ and $$B$$ is 5, $$A$$ will win. 