# SOLUTION: 2. Given 20 different types of inputs to a program, in how many ways can 8 of them be selected if a) Order does not matter b) Order matters

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 Click here to see ALL problems on Probability-and-statistics Question 549345: 2. Given 20 different types of inputs to a program, in how many ways can 8 of them be selected if a) Order does not matter b) Order matters Answer by Edwin McCravy(8909)   (Show Source): You can put this solution on YOUR website!2. Given 20 different types of inputs to a program, in how many ways can 8 of them be selected if a) Order does not matter b) Order matters ```Let's do the b) part first, where order matters: b) 1. There are 20 ways to select the 1st input. 2. For each of those 20 ways to select the 1st input, there are 19 ways left to select the 2nd input. That's 20×19 ways to select the first 2 inputs. 3. For each of those 20×19 ways to select the 1st 2 inputs, there are 18 ways to select the 3rd input. That's 20×19×18 ways to select the first 3 inputs. 4. For each of those 20×19×18 ways to select the 1st 3 inputs, there are 17 ways left to select the 4th input. That's 20×19×18×17 ways to select the first 4 inputs. 5. For each of those 20×19×18×17 ways to select the 1st 4 inputs, there are 16 ways left to select the 5th input. That's 20×19×18×17×16 ways to select the first 5 inputs. 6. For each of those 20×19×18×17×16 ways to select the 1st 5 inputs, there are 15 ways left to select the 6th input. That's 20×19×18×17×16×15 ways to select the first 6 inputs. 7. For each of those 20×19×18×17×16×15 ways to select the 1st 6 inputs, there are 14 ways left to select the 7th input. That's 20×19×18×17×16×15×14 ways to select the first 7 inputs. 8. For each of those 20×19×18×17×16×15×14 ways to select the 1st 7 inputs, there are 13 ways left to select the 8th input. That's 20×19×18×17×16×15×14×13 ways to select the first 8 inputs. That's all of them! So the answer is 20×19×18×17×16×15×14×13 = 5079110400 ----------------- Now let's do the a) part, where order does not matter. Each group of 8 in which the order does not matter was ordered in 8×7×6×5×4×3×2×1 ways above among the 20×19×18×17×16×15×14×13 in which order mattered. Therefore the answer to b) counts each group of 8 in the a) part 8×7×6×5×4×3×2×1 times too many. So we just need to divide the answer to part b) by 8×7×6×5×4×3×2×1 to get the answer for part a), and when we do we get: = = 125970 The short way is, of course. (a) 20C8 = 125970 (b) 20P8 = 5079110400 The first way is the reason WHY you do what you do. It may not seem so, but mathematics is easier in the long run when you understand WHY you do what you do, not just HOW to do it. Edwin```