Permutations/combinations problem

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However, you can simplify the equation by using the binomial theorem to rewrite it as p=(\lambda-1)^n-\binom{n}{0}+(\lambda-1)^{n-1}-\binom{n}{1}+(\lambda-1)^{n-2}-\binom{n}{2}+...+(\lambda-1)^{n-k}-\binom{n}{k}. This may make the equation look more complicated, but it is a more compact way of expressing the sum. In summary, when calculating the number of possible vectors with a certain restriction on the number of non-zeros, the equation is p=\sum_{k'=1}^{k}\left[(\lambda-1)^{k
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JL3
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I have a problem about combinations and permutations I am trying to solve. Say we have an n-dimensional vector. Each element of the vector can contain anyone of [tex]\lambda=3[/tex] values (-1, 0 or +1). Then the number of possible vectors is simply:

[tex]\lambda^n[/tex]

If we place the additional restriction that the vector must contain exactly [tex]k[/tex] non-zeros, then it becomes:

[tex]p=(\lambda-1)^{k}\times\binom{n}{k}=\frac{n!(\lambda-1)^{k}}{k!(n-k)!}[/tex]

If we change the restriction so that it must contain at most [tex]k[/tex] non-zeros and at least 1 non-zero, then it becomes:

[tex]p=\sum_{k'=1}^{k}\left[(\lambda-1)^{k'}\times\binom{n}{k'}\right]=\sum_{k'=1}^{k}\frac{n!(\lambda-1)^{k'}}{k'!(n-k')!}[/tex]

Are my equations correct? Is there a more compact way of expressing this last equation, to get rid of the summation?
 
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  • #2
Yes, your equations are correct. Unfortunately, there is no more compact way of expressing the last equation as it involves a sum of terms.
 

What is the difference between permutations and combinations?

Permutations are arrangements of items in a specific order, while combinations are selections of items without regard to order.

How do I calculate the number of permutations?

The number of permutations can be calculated by using the formula n! / (n-r)! where n is the total number of items and r is the number of items being selected.

How do I calculate the number of combinations?

The number of combinations can be calculated by using the formula n! / (r! * (n-r)!) where n is the total number of items and r is the number of items being selected.

What is the difference between with and without replacement in permutations/combinations?

With replacement means an item can be chosen more than once, while without replacement means an item can only be chosen once.

How do I approach a permutation/combinations problem?

First, identify if the problem involves permutations or combinations. Then, determine if the items are being selected with or without replacement. Finally, use the appropriate formula to calculate the number of possible arrangements or selections.

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