My Math Forum k - permutation and indices...

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October 7th, 2017, 11:32 PM   #1
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k - permutation and indices...

I am reading a russian textbook about Combinatorics. In the very beginning of the textbook, the author explains k-permutations. He considers a set of X (the first expression in the attached file). He defines every possible k-permutation of this set with the second expression in the attached file.

Although I understand the concept of permutation... I can't understand the meaning of the indices (i1, i2... ik) in the second expression. How they correlate with the indices in the first expression? The author doesn't give any explanation about the indices.I know it is all about "re-arrangement", but how is this rearrangement expressed by the i-indices?

I hope I put my question clearly enough, If not- i would try to make it more clear. I will be thankful for every comment on the topic!
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 October 8th, 2017, 03:38 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 The original order is $\displaystyle x_1,x_2, ..., x_n$. The permutation is $\displaystyle x_{i1}, x_{i2}, ..., x_{in}$. Here, "i1" is the original index of the "x" that has been moved to the first place, "i2" is the original index of the "x" that has moved to the second place, etc. For example, if $\displaystyle x_1, x_2, x_3, x_4$ has been permuted to $\displaystyle x_3, x_1, x_4, x_2$ then i1= 3, i2= 1, i3= 4, and i4= 2.
October 8th, 2017, 04:15 AM   #3
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Quote:
 Originally Posted by Country Boy if $\displaystyle x_1, x_2, x_3, x_4$ has been permuted to $\displaystyle x_3, x_1, x_4, x_2$ then i1= 3, i2= 1, i3= 4, and i4= 2.
I understand the example..... But how does it come that 3 is ascribed to i1, 1 is ascribed to i2 and so on... (in your example)? How is it defined?
It is defined by the concrete permutation.

If we consider another permutation, then: i1, i2, i3.. will have different values.
This is what I can't grasp.....

It turns out i1, i2... are indices, which are defined by the context (i mean by the concrete permutation)....

 October 8th, 2017, 06:17 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 I am not sure what you are asking. "3 is ascribed to i1, 1 is ascribed to i2 and so on" because that happens to be true for this particular permutation! There are 4!= 24 different permutations of four objects (in general there are n! different permutations of n objects). Since I don't want to write 24 different permutations, I will give an example with 3 objects, $\displaystyle x_1$, $\displaystyle x_2$, and $\displaystyle x_3$. Those 3!= 6 permutations are $\displaystyle \{x_1, x_2, x_3\}$ $\displaystyle \{x_1, x_3, x_4\}$ $\displaystyle \{x_2, x_1, x_3\}$ $\displaystyle \{x_2, x_3, x_4\}$ $\displaystyle \{x_3, x_1, x_2\}$ $\displaystyle \{x_3, x_2, x_1\}$ In the first, $\displaystyle i1= 1$, $\displaystyle i2= 2$, $\displaystyle i3= 3$. In the second, $\displaystyle i1= 1$, $\displaystyle i2= 3$, $\displaystyle i3= 2$. In the third, $\displaystyle i1= 2$, $\displaystyle i2= 1$, $\displaystyle i3= 3$. In the fourth, $\displaystyle i1= 2$, $\displaystyle i2= 3$, $\displaystyle i3= 1$. In the fifth, $\displaystyle i1= 3$, $\displaystyle i2= 1$, $\displaystyle i3= 2$. In the sixth, $\displaystyle i1= 3$, $\displaystyle i2= 2$, $\displaystyle i3= 1$.

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