Find the number of permutations of 'n' objects taken n - 1 at any time for any positve interger 'n'. Compare this answer with the number of permutations of all 'n' objects. Does this make sense? Explain.
2007-01-20 14:50:32 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
There are n! permutations of n objects. In addition there are n! permutations of n-1 of them. There are n ways to pick the object not in the permutation, and then (n-1)! ways to permute the remaining n-1 objects.
For example, take the 3 objects a,b,c. The permuations of all three are:
abc
acb
bac
bca
cab
cba
The permuations of two of them are:
ab
ac
ba
bc
ca
cb
Six of each!
2007-01-20 15:13:57 · answer #1 · answered by Phineas Bogg 6 · 0⤊ 0⤋
a permutation is like a lottery drawing where no two numbers are the same. For example lets say there is a lottery where you choose 6 numbers from 1 to 39 but you can't pick the same number twice. So to calculate the total number of permutations where there are six objects with n possibilities in object 1 and n - 1 in all following objects where n = the possibilities of the previous object, and multiply all possibilities of all six objects.
n1 = 39
n2 = 38
n3 = 37
n4 = 36
n5 = 35
n6 = 34
total permutations 39 * 38 * 37 * 36 * 35 * 34 = 2,349,088,560
2007-01-20 23:57:23 · answer #2 · answered by ikeman32 6 · 0⤊ 0⤋
They are exactly the same. You have n x (n-1)! in the first case, and n! in the second case.
2007-01-20 23:39:53 · answer #3 · answered by sahsjing 7 · 0⤊ 0⤋