Nope, that is the combination of {x,y,z} taking 2 objects at a time, because combination has no regard for specific order. The permutations for {x,y,z} taking two at a time are: xy, xz, yx, yz, zx, zy.
2006-06-25 15:16:50
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answer #1
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answered by Samara 2
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If you consider all two letter words containing x,y and / or z never double, then
The permutations of {x,y,z} are xy, yz, and xz and no others
is false,
because there is as well yx zx and zy
If you consider all products containing x,y and / or z never double, then
The permutations of {x,y,z} are xy, yz, and xz and no others
is true
2006-06-25 15:59:19
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answer #2
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answered by Thermo 6
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False.
The order matters in permutation and also the number of items
So fir 3 different items x y and z one at a time there are 3 permutations written as 3P1. i.e. x,y,z. This could be considered a trivial case
For 2 at a time it is 3P2 or 6 i.e. xy, yz, zx, yx,zy,xz
for 3 at a time 3P3 or 3! or 6 i.e. xyz, xzy, yxz,yzx, zxy, zyx
2006-06-25 15:40:00
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answer #3
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answered by wvl 3
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False.
Those three are permutations of {x,y,z} but you forgot:
x
y
z
yx
zy
zx
xyz
xzy
yxz
zxy
yzx
zyx
Permutations are groups of elements in which the order is important. You also didn't mention the number of elements in each permutation, so it can be 1,2 or 3.
Hope this helps^_^
2006-06-25 23:17:56
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answer #4
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answered by kevin! 5
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xy, yz, and xz are combinations of x,y,z
xy, yx, yz, zy, xz, and zx are permutations of x,y,z
2006-06-25 17:07:30
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answer #5
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answered by Sherman81 6
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F - those are combinations.
permutations are xyz yzx ... - all possible ways to re-order the elements
2006-06-25 14:59:30
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answer #6
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answered by Anonymous
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zx andzy
2006-06-25 15:00:32
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answer #7
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answered by nettie l 1
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I believe the empty set { } is one also.
2006-06-25 15:00:44
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answer #8
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answered by peace_n_luv 3
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False....x, y, z, xyz
2006-06-25 15:00:27
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answer #9
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answered by johndersel 3
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permutations are -
{}, {x},{y},{z}, {x,y}, {y,z} , {z,x}, {y,x}, {z,y} , {x,z},{x,y,z}, {y,z,x}, {z,x,y },{x,z,y} ,{y,x,z} , {z,y,x}
2006-06-25 16:22:49
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answer #10
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answered by Anonymous
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