Answer to Question #98563 in Combinatorics | Number Theory for musa

Question

In how many ways can ball bearings labeled A, B, C and D be
shared by Jacob and Helen in such a way that
1 Jacob gets one ball bearing and Helen gets three
2 They each get two ball bearings?

Answer 1

1

$$A+BCD, B+ACD, C+ABD, D+ABC$$

Four ball bearings labeled A, B, C and D can be shared by Jacob and Helen in such a way that Jacob gets one ball bearing and Helen gets three in four ways.

2

$$AB+CD,AC+BD, AD+BC,$$ $$BC+AD, BD+AC,CD+AB$$

$$\binom{4}{2}={4! \over 2!(4-2)!}={4\cdot3 \over 1\cdot2}=6$$

Four ball bearings labeled A, B, C and D can be shared by Jacob and Helen in such a way that Jacob gets two ball bearings and Helen gets two in six ways.