How many different ways(orders) can 5 people stand ?

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Not sure if you have Chad's videos but they help tremendously with determining between permutation and combinations. Chad simplifies it so that you don't necessarily have to remember the long and confusing formulas for the two. In a permutation, order matters. In a combination, order does not matter so in a combination you will have to divide by the number of selections factorial whereas you do not in a permutation. Combination Example: Twelve people have entered a raffle in which four will be selected to win. How many sets of four winners are possible? Here, order does not matter. Your answer would be (12x11x10x9)/(4x3x2x1). Does this make sense? We have 12 selections for the first winner, 11 for the second, 10 for the third, 9 for the fourth and because order doesn't matter we need to divide by the factorial of # of selections which in this case is 4 (4 winners are picked.)
So for #14 in Destroyer... Order does matter, it is a permutation! So the answer will simply be 5 factorial which = 120.
 
Not sure if you have Chad's videos but they help tremendously with determining between permutation and combinations. Chad simplifies it so that you don't necessarily have to remember the long and confusing formulas for the two. In a permutation, order matters. In a combination, order does not matter so in a combination you will have to divide by the number of selections factorial whereas you do not in a permutation. Combination Example: Twelve people have entered a raffle in which four will be selected to win. How many sets of four winners are possible? Here, order does not matter. Your answer would be (12x11x10x9)/(4x3x2x1). Does this make sense? We have 12 selections for the first winner, 11 for the second, 10 for the third, 9 for the fourth and because order doesn't matter we need to divide by the factorial of # of selections which in this case is 4 (4 winners are picked.)
So for #14 in Destroyer... Order does matter, it is a permutation! So the answer will simply be 5 factorial which = 120.
Ok so... When combination - we just do # factorial / # selection factorial ?
For permutation - we do just the # factorial

Is this true for all permutation and combination problems ? Since you said, we don't have to really memorize those formulas
 
The way Chad has explained it, yeah I believe that those are good explanations for how to solve the problems. Technically we are still using the long and complicated formula but Chad has simplified it down into understandable terms. 🙂
 
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