- #1
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 \lambda=3 values (-1, 0 or +1). Then the number of possible vectors is simply:
\lambda^n
If we place the additional restriction that the vector must contain exactly k non-zeros, then it becomes:
p=(\lambda-1)^{k}\times\binom{n}{k}=\frac{n!(\lambda-1)^{k}}{k!(n-k)!}
If we change the restriction so that it must contain at most k non-zeros and at least 1 non-zero, then it becomes:
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')!}
Are my equations correct? Is there a more compact way of expressing this last equation, to get rid of the summation?
\lambda^n
If we place the additional restriction that the vector must contain exactly k non-zeros, then it becomes:
p=(\lambda-1)^{k}\times\binom{n}{k}=\frac{n!(\lambda-1)^{k}}{k!(n-k)!}
If we change the restriction so that it must contain at most k non-zeros and at least 1 non-zero, then it becomes:
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')!}
Are my equations correct? Is there a more compact way of expressing this last equation, to get rid of the summation?
