SOLUTION: 5^C2 + 5^C1

Algebra ->  Permutations -> SOLUTION: 5^C2 + 5^C1      Log On


   

Question 251519: 5^C2 + 5^C1
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
n%21%2F%28n-r%29%21r%21 Start with the given formula



5%21%2F%285-2%29%212%21 Plug in n=5 and r=2



5%21%2F3%212%21 Subtract 5-2 to get 3


Expand 5!
%285%2A4%2A3%2A2%2A1%29%2F3%212%21


Expand 3!
%285%2A4%2A3%2A2%2A1%29%2F%283%2A2%2A1%292%21



%285%2A4%2Across%283%2A2%2A1%29%29%2F%28cross%283%2A2%2A1%29%292%21 Cancel



%285%2A4%29%2F2%21 Simplify


Expand 2!
%285%2A4%29%2F%282%2A1%29



20%2F%282%2A1%29 Multiply 5*4 to get 20



20%2F2 Multiply 2*1 to get 2



10 Now divide



So 5 choose 2 (where order doesn't matter) yields 10 unique combinations

---------------------------------------------------------------------------



n%21%2F%28n-r%29%21r%21 Start with the given formula



5%21%2F%285-1%29%211%21 Plug in n=5 and r=1



5%21%2F4%211%21 Subtract 5-1 to get 4


Expand 5!
%285%2A4%2A3%2A2%2A1%29%2F4%211%21


Expand 4!
%285%2A4%2A3%2A2%2A1%29%2F%284%2A3%2A2%2A1%291%21



%285%2Across%284%2A3%2A2%2A1%29%29%2F%28cross%284%2A3%2A2%2A1%29%291%21 Cancel



%285%29%2F1%21 Simplify


Expand 1!
%285%29%2F%281%29



5%2F%281%29 Multiply 5 to get 5



5%2F1 Multiply 1 to get 1



5 Now divide



So 5 choose 1 (where order doesn't matter) yields 5 unique combinations



Since and , this means