boloman0z
2022-06-26
Answered

Is it possible to quantify the number of dimensions in combinatorial spaces. The space I am particularly interested is all partitions of a set, bounded by the Bell number, where objects in this space are particular partitions.

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Blaine Foster

Answered 2022-06-27
Author has **33** answers

It makes sense to consider some sets of combinatorial objects as spaces (or polytopes) and therefore discuss dimensionality (e.g. the set of n by n (-1,0,+1)-matrices). Although, perhaps the word "dimension" could better be described as "degrees of freedom".

Mathematically, degrees of freedom is the dimension of the domain of a random vector, or essentially the number of 'free' components: how many components need to be known before the vector is fully determined.

I suspect that it will be difficult to discuss dimensionality in many combinatorial settings. For example, imagine constructing a Latin square, starting from an empty matrix, placing symbols one-at-a-time in a non-clashing manner. After placing (say) half of the symbols, we might find: (a) there are still many completions of this partial Latin square, (b) there are no completions of this partial Latin square or (c) there is a unique completion of this partial Latin square. This seems to go against the notion of dimensionality -- the number of "components" required to determine the Latin square is not fixed.

You could think of the set of partitions of a set of n elements as having dimension n. You require n pieces of information to determine the partition. There's no point in having a notion of "dimension" unless you can use it for something.

Mathematically, degrees of freedom is the dimension of the domain of a random vector, or essentially the number of 'free' components: how many components need to be known before the vector is fully determined.

I suspect that it will be difficult to discuss dimensionality in many combinatorial settings. For example, imagine constructing a Latin square, starting from an empty matrix, placing symbols one-at-a-time in a non-clashing manner. After placing (say) half of the symbols, we might find: (a) there are still many completions of this partial Latin square, (b) there are no completions of this partial Latin square or (c) there is a unique completion of this partial Latin square. This seems to go against the notion of dimensionality -- the number of "components" required to determine the Latin square is not fixed.

You could think of the set of partitions of a set of n elements as having dimension n. You require n pieces of information to determine the partition. There's no point in having a notion of "dimension" unless you can use it for something.

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Combinatorial proof of a Fibonacci identity: $n{F}_{1}+(n-1){F}_{2}+\cdots +{F}_{n}={F}_{n+4}-n-3.$

Does anyone know a combinatorial proof of the following identity, where ${F}_{n}$ is the $n$th Fibonacci number?

Does anyone know a combinatorial proof of the following identity, where ${F}_{n}$ is the $n$th Fibonacci number?

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Simple combinatorics and probability theory related question

5 apples are randomly distributed to 4 boxes. We need to find probability that there are 2 boxes with 2 apples, 1 box with 1 apple and 1 empty box.

I'm getting the correct answer with $\frac{\frac{5!}{2!2!1!0!}\ast 4\ast 3}{{4}^{5}}=0.3515625$ (anyway, the answer is said to be 0.35, but I think it is a matter of rounding).

But I don't understand why there are ${4}^{5}$ elementary events in total. Firstly, I thought It should be $(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{4}{5}{\textstyle )}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}})$ - number of combinations with repetitions, but I couldn't get the proper answer.

Isn't approach with $(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{4}{5}{\textstyle )}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}})$ elementary events more correct?

5 apples are randomly distributed to 4 boxes. We need to find probability that there are 2 boxes with 2 apples, 1 box with 1 apple and 1 empty box.

I'm getting the correct answer with $\frac{\frac{5!}{2!2!1!0!}\ast 4\ast 3}{{4}^{5}}=0.3515625$ (anyway, the answer is said to be 0.35, but I think it is a matter of rounding).

But I don't understand why there are ${4}^{5}$ elementary events in total. Firstly, I thought It should be $(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{4}{5}{\textstyle )}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}})$ - number of combinations with repetitions, but I couldn't get the proper answer.

Isn't approach with $(\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{4}{5}{\textstyle )}\phantom{\rule{negativethinmathspace}{0ex}}\phantom{\rule{negativethinmathspace}{0ex}})$ elementary events more correct?

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given a multiset (a set with repetitions allowed) of $2n+1$ real numbers, say $S=\{{r}_{1},\dots ,{r}_{2n+1}\}$.

These numbers are such that for every $k$, the multiset $S-\{{r}_{k}\}$ can be split into two multisets of size $n$ each, such that the sum of numbers in one multiset is same as the sum of numbers in the other.

Show that all the numbers must be equal.( i.e. ${r}_{i}={r}_{j}$)

These numbers are such that for every $k$, the multiset $S-\{{r}_{k}\}$ can be split into two multisets of size $n$ each, such that the sum of numbers in one multiset is same as the sum of numbers in the other.

Show that all the numbers must be equal.( i.e. ${r}_{i}={r}_{j}$)

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b. What is the probability of finding the winner on the 10th call?

c. What is the probability that the station will need 9 or more calls to find a winner?

a. What is the PMF (Probability Mass Function) of L, the number of calls necessary to find the winner?

b. What is the probability of finding the winner on the 10th call?

c. What is the probability that the station will need 9 or more calls to find a winner?