If the police have 9 suspects how many different ways can they select 5 for a lineup

Assuming the order does not matter, you want the number of combinations of 9 things taken 5 at a time.  The combinations can be shown as C(9,5), 9C5.

C(9, 5) =
9/5(9-5) = 
9*8*7*6*5 / 5*4 

The 5 terms cancel. 

9*8*7*6 / 4*3*2 = 
9*7*2 = 
126

The above change is because 4*2 cancels the 8 in the numerator and 6/3 = 2
Therefore, the solution is 126. 

The third question is correct.

If i’m not wrong i believe it is c