Permutations and combinations? please help again?

ok i got that now i'm not sure for these answers-





- how many sandwichess are possible if the restuarant lets you build a sandwicj by chosing any 4 of 10 sandwich toppings?





- if there are 6 soups to chose from , how many soup- and build a sandwich specials are there?





- an inline skate has 4 wheels . how many ways could 4 replacement wheels be chosen from a pack o 10 wheels and fitted to a skate?





not sure if they are permutatons or combinations please help

Permutations and combinations? please help again?
note


if ham on top of sausage is different than


sausage on top of ham


then this is a permutation


but if the order does not matter then it is a combination


(i.e. ham then sausage, sausage then ham)


so


nPr = n!/(n-r)!





10P6 = 10!/(10-4)! = 10!/6! = [10*9*8*7*6!) / (6!) =


10(9)(8)(7)
Reply:The sandwiches are combinations, unless someone cares which topping goes on top of which other topping (got that?). Assuming the order doesn't matter, it's


10C4 = 10! / [(10-4)! * 4!]


= 10 * 9 * 8 * 7 / (4 * 3 * 2) = 210





The choices of soup and sandwich are independent, so the soup and sandwich combos are


6 * 210 = 1260 choices.





Does it really matter which wheel comes out of a new pack? If so, does it matter which position on the skate each wheel goes on? If it doesn't, it's 10C4 = 210 again.





If the position on the skate matters, then we need to do permutations instead. But the permutations are just 4! times the number of combinations, so multiply:


210 * 24 = 21024



sweating