Itemized Statement Sample: Fill & Download for Free

A simple yet effective approach is to track research tasks and resources with a spreadsheet, especially an online one that allows for collaboration.Here's one way you could structure the different items:Category | Item | Priority | Topic | Tags | Link | DescriptionwhereCategory = question, reference, article, finding, etcItem = statement of the question, title of the paper, etcPriority = 1-5Topic = research subjectTags = Keywords, maybe comma-separatedLink = URL or file pathDescription = any other detailsYou can use filtering to view items with a specific category say, or with a given tag.I've posted a sample spreadsheet with this structure at Template - Research Project - Google Sheets - feel free to comment on it and to reuse.

When you pay for a cup of tea in a five star hotel, you are not paying for just a cup of fragrant hot water. You are paying for:The rent for the land over which the restaurant is builtThe fancy cutlery your tea is served inThe furniture and everything else that contributes to the five-star ambienceThe energy cost for the air-conditioning and lightingThe salary of the waiting staff and the chefsOther miscellaneous expenses that are part of the Income Statement of any business.The COGS (Cost of Goods Sold) in restaurants is usually around 30%. For high margin items such as beverages, they are likely to be even lower.Here's a sample income statement (also called Profit and Loss Statement for a restaurant I found on the internet:See how COGS is roughly 30% of the revenues. If you read the various rows under Operating Expenses you will get an idea of what all costs go into running any such establishment. All those costs will ultimately have to be recovered from YOU.A road-side tea vendor, on the other hand, might incur similar kinds of costs, but they are much smaller. Also, a five-star hotel usually has a higher commitment to quality, both for ingredients and for service, and all that costs even more money.Later edit:It should also be borne in mind that a 5-star hotel is essentially a luxury offering, and, to a certain extent, the price helps create and maintain the perception of luxury.

Thinking generically we model people with objects or items and we model groups (of people) with buckets or boxes.With the above transition in hand the vague objective then is to describe mathematically the distribution of objects across buckets.Objects can be either anonymous (or indistinguishable) or they can be distinct. And so are the buckets.Since we are not told explicitly which ones are which, we shall take it here that all the objects or all the items are distinct and we shall consider the distinct buckets first and indistinguishable buckets second.It may seem that we are now ready to tighten or formalize our requirement as follows:find the number of ways to distribute [math]n[/math] distinct object across [math]k[/math] distinct bucketsbut there is a reason why combinatorics is often labelled as a low constraint discipline since we have to be very precise in our description of the problem statement because the unfolding solution path is very sensitive to such a statement.We thus postulate the following requirements:the distribution rule must be given ahead of timeeach item must wind up in some bucketmultiple items per bucket are allowedthe order of items within the bucket is irrelevantthe concept of order of items across the buckets does not applyAssume that we have an input set of [math]n = 5[/math] (distinct) items:[math]S = \left\{A, B, C, D, E\right\} \tag*{}[/math]and [math]k = 2[/math] distinct buckets denoted [math]b_1[/math] and [math]b_2[/math].The sample distribution rule - place [math]r_1 = 2[/math] items into [math]b_1[/math] and [math]r_2 = 3[/math] items into [math]b_2[/math] - ensures that, as required, it is the case that [math]2+3=5[/math] meaning that [math]r_1+r_2 = n[/math] and [math]k[/math] is equal to the number of the summands [math]r_j[/math].The first event of placing [math]2[/math] items into [math]b_1[/math], combinatorially, amounts to the selection of [math]2[/math] distinct items from the initial pool of [math]5[/math] items which can be carried out in[math]\displaystyle \binom{5}{2} = \dfrac{5!}{2!(5-2)!} \tag{1}[/math]ways:[math]AB \tag*{}[/math][math]AC \tag*{}[/math][math]AD \tag*{}[/math][math]AE \tag*{}[/math][math]BC \tag*{}[/math][math]BD \tag*{}[/math][math]BE \tag*{}[/math][math]CD \tag*{}[/math][math]CE \tag*{}[/math][math]DE \tag*{}[/math]Note that as promised (or as required) we are ignoring the internal order of the selected items inside the bucket [math]b_1[/math].Once the above [math]2[/math] items have been selected, there will be [math](5 - 2)[/math] items left in the pool.The second event of placing [math]3[/math] items into [math]b_2[/math], combinatorially, amounts to the selection of [math]3[/math] items from the remaining [math](5 -2)[/math] which can be carried out in[math]\displaystyle \binom{(5-2)}{3} = \dfrac{(5-2)!}{3!(5-2-3)!} \tag{2}[/math]ways:[math]CDE \tag*{}[/math][math]BDE \tag*{}[/math][math]BCE \tag*{}[/math][math]BCD \tag*{}[/math][math]ADE \tag*{}[/math][math]ACE \tag*{}[/math][math]ACD \tag*{}[/math][math]ABE \tag*{}[/math][math]ABD \tag*{}[/math][math]ABC \tag*{}[/math]Note that again as promised (or as required) we are ignoring the internal order of the selected items inside the bucket [math]b_2[/math].Observe that combinatorially the first and the second events are independent of each other. There is a connection between these two events in the sense that the selection of the concrete items for the first event removes these items from the pool and, thus, makes them unavailable for the consequent choices.But we are dealing here with the items available in a fixed (and, thus, finite) supply - the removal of the items after the first event is not compensated with similar items taken from elsewhere.Since the Multiplication Counting Principle (MCP) is only concerned about the number - not the content - once the numbers [math]r_1 = 2[/math] and [math]r_2 = 3[/math] are fixed ahead of time, the outcome of the fist event does not affect the number of outcomes of the second event.Thus, by MCP the above two events can occur in sequence in a total of[math]\dfrac{5!}{2!(5-2)!}\times \dfrac{(5-2)!}{3!(5-2-3)!} \tag*{}[/math]ways. Here we did not do any actual computations on purpose - to show that the common term [math](5 - 2)![/math] will cancel out. If we denote the answer as [math]P(5, 2, 3)[/math] then:[math]P(5, 2, 3) = \dfrac{5!}{2!\cdot 3!\cdot 1} \tag*{}[/math]because[math](5 - 2 - 3)! = (5 - (2 + 3))! = (5 - 5)! = 0! = 1 \tag*{}[/math]The above argument can be generalized in a straightforward manner:if [math]n[/math] distinct objects are distributed across [math]k[/math] distinct buckets [math]b_1, \ldots , b_k[/math] in such a way that [math]r_1[/math] items are placed into [math]b_1[/math], [math]r_2[/math] items are placed into [math]b_2[/math] and so on until [math]r_k[/math] items are placed into [math]b_k[/math] and[math]\displaystyle n = \sum_{j=1}^kr_j \tag{3}[/math]then the total number of such distributions [math]P\left(n, r_1, \ldots, r_k\right)[/math] is[math]P\left(n, r_1, \ldots, r_k\right) = \dfrac{n!}{r_1!\cdot r_2!\cdot\ldots\cdot r_k!} \tag{4}[/math]which is also known as the multinomial coefficient.The proof of the above statement essentially amounts to the recitation of our previous reasoning in generic terms: there are a total of[math]\displaystyle \binom{n}{r_1} = \dfrac{n!}{r_1!\cdot\left(n-r_1\right)!} \tag*{}[/math]ways to choose [math]r_1[/math] distinct items from [math]n[/math], there are a total of[math]\displaystyle \binom{n-r_1}{r_2} = \dfrac{\left(n-r_1\right)!}{r_2!\cdot\left(n-r_1-r_2\right)!} \tag*{}[/math]ways to choose [math]r_2[/math] distinct items from [math]n-r_1[/math], there are a total of[math]\displaystyle \binom{n-r_1-r_2}{r_3} = \dfrac{\left(n-r_1-r_2\right)!}{r_3!\cdot\left(n-r_1-r_2-r_3\right)!} \tag*{}[/math]ways to choose [math]r_3[/math] distinct items from [math]n-r_1-r_2[/math] and so on.The total number of ways to place [math]r_1[/math] distinct items into [math]b_1[/math] AND [math]r_2[/math] distinct items into [math]b_2[/math] AND [math]r_3[/math] distinct items into [math]b_3[/math] and so on, by MCP, is[math]\dfrac{n!}{r_1!\cdot\left(n-r_1\right)!}\cdot\dfrac{\left(n-r_1\right)!}{r_2!\cdot\left(n-r_1-r_2\right)!}\cdot\dfrac{\left(n-r_1-r_2\right)!}{r_3!\cdot\left(n-r_1-r_2-r_3\right)!} \cdot \tag*{}[/math][math]\dfrac{\left(n-r_1-r_2-\ldots-r_{k-1}\right)!}{r_k!\cdot\left(n-r_1-r_2-r_3-\ldots -r_k\right)!} \cdot \tag*{}[/math]In our rendering the compound factorials of the shape [math](n-r_1-r_2)![/math] and such are present in both the denominator and the numerator of all the fractions and will, thus, diagonally, cancel out:The compound factorial in the denominator of the last fraction amounts to a unity:[math]\displaystyle \left(n - r_1 - r_2 - \ldots - r_k \right)! = \left(n - \left(\sum_{j=1}^kr_j\right)\right)! = 0! = 1 \tag*{}[/math]as per (3).Consequently, if the buckets are indistinguishable then by the Division Counting Principle we reduce the answer in (4) by [math]k![/math]:[math]\dfrac{P\left(n, r_1, \ldots, r_k\right)}{k!} = \dfrac{n!}{r_1!\cdot r_2!\cdot\ldots\cdot r_k!\cdot k!} \tag{5}[/math]since [math]k![/math] is the number of linear permutations of [math]k[/math] distinct items.In your case we have especially convenient numbers - we seek the number of ways to distribute [math]m\cdot n[/math] distinct items across [math]m[/math] distinct buckets where exactly [math]n[/math] items are placed into each bucket.From (4) the answer in that case is:[math]\dfrac{(m\cdot n)!}{\left(n!\right)^m} \tag{6}[/math]because in this case each [math]r_j = n[/math] and there are exactly [math]m[/math] of them:[math]\underbrace{n!\cdot n!\cdot n!\cdot\ldots\cdot n!}_{m} = \left(n!\right)^m \tag*{}[/math]If, however, the groups of people are indistinguishable (which in my humble and subjective opinion is, in vacuum, more likely to be the case) then the answer is:[math]\dfrac{(m\cdot n)!}{\left(n!\right)^m\cdot m!} \tag{7}[/math]Note that from the problem-solving perspective we solved a more general problem first and then descended, in a way, to a simpler or a specific case.To do that we used the following approach - we took a concrete problem and abstracted it away. Abstraction can be thought of as a mechanism of separation of irrelevant minutiae from the essence of the phenomenon at hand.Abstraction is a concept that is used much not only in (of course) mathematics but also in computer science, computer programming and physics. It turns out that we live in a rather peculiar universe - it often happens that after we come up with an abstraction based on just one specific case, later on there pop into existence other, seemingly totally disconnected domains of knowledge, for which our abstraction can be used as a template.For example. Here we switched from people and groups to items and buckets.Splendid. Now what?To demonstrate the deep connection between mathematics and physics - a point of frequent inquiries here on Quora - consider the three popular models known as Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac statistics which study the behavior of systems comprised of a large number of weakly interacting particles in thermal equilibrium.In all three statistical models the role of items is played by the idealized particles and the role of buckets is played by the amount of energy these particles possess. The energy levels are taken to always be distinct while the particles may or may not be distinct.All three statistics aspire to answer the following question:if some particle is selected at random then what is the probability that it will have a specific (allowed) amount of energy?In Maxwell-Boltzmann statistics the particles (gas molecules) are classical and thus possess the fundamental property that we call a trajectory - some curve, [math]\vec{r}(t)[/math], which is representable as a real-valued function of a real-valued argument at least twice differentiable. We hope that we can recover such a function by solving the equations of Newton/Lagrange/Hamilton.The particles in MB-statistics are taken to be distinct in the sense that we can number them and track them over time. It follows then that in MB-statistics we are dealing with the distributions of distinct items across distinct buckets. Here multiple particles are allowed to have the same energy level - be in the same bucket.This is neither the time nor the place for all the technical details but, in general, to find the number of all the possible distributions of items in this case we would combine the number of possible states across the buckets given by our number [math]P\left(n, r_1, \ldots, r_k\right)[/math] with the number of possible states within a single bucket given by [math]n^r[/math].In quantum mechanics things get very strange very fast and the concept of trajectory flies out the window.In Bose-Einstein statistics multiple particles (photons, atomic nuclei, atoms with an even number of elementary particles or, collectively, bosons) are allowed to have the same energy level - be in the same bucket. Since these particles are indistinguishable, what matters is not what kind of particles are in any given bucket but rather how many of them are in it. The number of possible arrangements of this type is given by the number of combinations of a multiset with infinite supply:[math]\dfrac{(n+r-1)!}{r!(n-1)!} \tag*{}[/math]In Fermi-Dirac statistics multiple particles (electrons, neutrons, protons or, collectively, fermions) are not allowed to occupy the same energy level. Here any given bucket can have no more than one particle. This is known as the Pauli Restriction or Pauli Exclusion.As such, at any given time the bucket is either empty or contains exactly one item. It follows then that in this case the number of possible distributions is equal to the number of ways in which it is possible to distribute [math]r[/math] indistinguishable items across [math]n[/math] distinct buckets given the binomial coefficient:[math]\displaystyle \binom{n}{r} \tag*{}[/math]In all three statistics the resulting numbers contain a factorial of the total number of particles and that factorial, as you might imagine, will normally be enormous. And if the factorial is enormous then its Sterling's approximation works well - that is how in broad strokes (overlooking some number of gory details) the relevant formulas are deduced.