Question 419824
sorry you got lead astray...


1.  there are 10C3 ways to select 3 dogs from 10
___ there are 15C4 ways to select 4 cats from 15
___ the number of ways to form the committee is the product of the possible numbers of the two groups
___ 10C3 multiplied by 15C4



2.  you have to start with ownership of all three and eliminate duplicate ownership
___ 3 owned c&t , but 2 of those owned all three , so 1 owned only c&t
___ 7 owned d&t , but 2 of those owned all three , so 5 owned only d&t
___ 12 owned d&c , but 2 of those owned all three , so 10 owned only d&c
___ 11 owned t , but 8 owned other pets , so 3 owned only t
___ 28 owned c , but 13 owned other pets , so 15 owned only c
___ 24 owned d , but 17 owned other pets , so 7 owned only d
___ 43 owned one or more pets , so 8 owned none



3.  C is 3 over and 2 down from A , so there are 5 steps to get there
___ you can take the steps in any order , but the 3 overs are the same and the 2 downs are the same
___ the number of different routes from A to C is ___ 5! / [(3!)(2!)]