Arrangment Problem Help

G

Guest

Guest
How many arrangements of six 0s, five 1s, and four 2s are there in which

a) The first 0 precedes the first 1?

For this one I figure it's something like this: 15!/((6-x)!(5-y)!4!) where x and y are used to show my confusion

b) The first 0 precedes the first 1, which precedes the first 2?

And for this one i have similar (most likely flawed) logic for
 
Problem b) is already solved, as daon pointed out. To solve problem a) you first put in the four 2's in four of the 15 possible positions. There are Binomial[15,4] ways to do that. Then in the first unoccupied position you put a 0. And in the remaining 10 positions you put in the remaining five 0's and five 1's. There are Binomial[10,5] ways to do that. So the answer is Binomial[15,4]* Binomial[10,5]
 
You must log in or register to reply here.
Top