How many ways are there to divide a red bracelet, a yellow bracelet, a green bracelet, and a black bracelet among 4 people if someone must receive exactly two bracelets?
+118608 Four people, 4 bracelets
One person gets exactly 2 so at least one person must get non.
2,1,1,0
4C2= 6 ways to choose the two
4 ways to chose the person who misses out
3 ways to chose the person to get 2,
2 ways to sort the rest
6*4*3*2 = 144 ways
2,2,0,0
6 ways to pair the braclets, (normally you would halve this but I am thinking differently.)
4C2 ways to choose two people =6
6*6 = 36 ways
180 ways total, I think....
