Question 1049716: Please do show me how did you solve it
Find the Number of Ways of Selecting 4 applications out of 12
There are seven entries in a horse race. In how many different ways can the horses finish?
Nine horses are running in a race. In how many different ways can those horses come in first, second, and third respectively?
Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Find the Number of Ways of Selecting 4 applications out of 12
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The 1st choice is 1 of 12.
Then 1 of 11, then 10, then 9.
12*11*10*9 = 11880
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But, if there's no difference in the applications, ie, if A, B, C & D is the same as B, C, D & A then there are 4*3*2*1 = 24 ways to make each group of 4.
11880/24 = 495 ways.
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There are seven entries in a horse race. In how many different ways can the horses finish?
7*6*5*4*3*2*1 = 5040 ways.
That's 7! called 7 factorial.
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Nine horses are running in a race. In how many different ways can those horses come in first, second, and third respectively?
9*8*7 ways.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! Find the Number of Ways of Selecting 4 applications out of 12.
the formula to use is c(n,x) = n! / (x! * (n-x)!)
this is because you want to know how many ways you can get x things out of n things where the arrangement of the things within each set is not important.
you will get c(12,4) = 12! / (4! * 8!) = 495.
There are seven entries in a horse race. In how many different ways can the horses finish?
you want to know how many different ways you can arrange a set of 7 things.
the formula for that is n!.
you will get 7! = 5040.
Nine horses are running in a race. In how many different ways can those horses come in first, second, and third respectively?
you want to know how many ways you can get x things out of n where order is important.
the formula to use is p(n,x) = n! / (n-x)!.
you will get p(9,3) = 9! / 6! = 504.
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what is the difference between order is important and order is not important?
a simple example may help explain the difference, and the relationship between the combination formula and the permutation formula.
assume a set contains the letters abc in that order.
if order is not important, then that is considered to be one set regardless of the possible arrangement of the letters in that set.
therefore, abc, acb, bac, bca, cab, cba are all considered to be members of that same set if order is not important, and are all considered to be members of different sets if order is important.
now suppose you have 3 horses in a race.
the horses are a, b, and c.
you want to find the possible ways you can get 2 out of those 3 horses to win the race if order is not important.
you would use the combination formula of c(n,x) = n! / (x! * (n-x)!)
the formula becomes c(3,2) = 3! / (2! * 1!) = (3*2*1) / (2*1*1) = 3.
there are 3 ways you can get 2 winners out of the race where order is not important.
those ways are ab, ac, bc.
each set contains a first place winners and a second place winner but you don't care which one.
order is not important.
now, if you do care about the arrangement within each of those sets, you multiply each of those sets by 2! to get 6 possible sets.
those sets are ab, ba, ac, ca, bc, cb.
it's the same horses, only now you care about which one is first and which one is second, therefore you consider each arrangement within each set as a different set.
the formula to use in that case would be p(n,x) = n! / (n-x)! which becomes p(3,2) = 3! / 1! which becomes (3*2*1) / 1) which becomes equal to 6.
from this you should be able to see the relationship between the combination formula and the permutation formula.
c(n,x) = p(n,x) / x!.
p(n,x) = c(n,x) * x!.
the number of possible ways you can get x things out of n where order is not important is c(n,x) = n! / (x! * (n-x)!).
multiply that by x! and you get the number of possible ways you can get x things out of n where order is important.
what you did was take each set of x things and then showed the number of possible ways you could arrange those x things.
that is, you multiplied n! / (x! * (n-x)!) by x! to get n! / (n-x)!, which is the permutation formula derived from the combination formula.
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