The first of these equalities is straightforward by the definition of binomial coefficients. To prove the second, we make use of partial fractions. We write the expansion

where the coefficients are to be determined. Multiplying both sides by for some ,

Then setting in both sides gives

leading to the desired result.

As a few examples of this identity we have:

I had previously posted an identity with similar, but trickier derivation here.

]]>- Most runs in at most n consecutive innings
- Most runs in at most n consecutive tests
- Most runs by date range (NEW)
- Highest average in n consecutive innings
- Most scores of at least m in at most n consecutive innings
- Most runs after n innings
- Highest average after n innings
- Most runs after n tests
- Highest average after n tests
- Innings to score at least m on k occasions
- Innings to score at least n runs
- Innings by innings batting list for a selected player

- Most wickets in at most n consecutive innings
- Most wickets in at most n consecutive tests
- Most wickets by date range (NEW)
- Lowest average in n consecutive innings
- Lowest economy rate in n consecutive innings
- Lowest strike rate in n consecutive innings
- Most m-wicket hauls in at most n consecutive innings
- Most wickets after n innings
- Lowest average after n innings
- Most wickets after bowling in n tests
- Lowest average after bowling in n tests
- Innings to take at least m wickets on k occasions
- Innings to take at least n wickets from debut
- Innings by innings bowling list for a selected player

- Most runs in at most n consecutive innings
- Most runs by date range (NEW)
- Highest average in n consecutive innings
- Highest strike rate in n consecutive innings
- Most scores of at least m in n consecutive innings
- Most runs after n innings
- Innings to score at least m on k occasions
- Innings to score at least n runs from debut
- Innings by innings batting list for a selected player

- Most wickets in at most n consecutive innings
- Most wickets by date range
- Lowest average in n consecutive innings
- Lowest economy rate in n consecutive innings
- Lowest strike rate in n consecutive innings
- Most m-wicket hauls in at least n consecutive innings
- Most wickets after n innings
- Lowest average after n innings
- Innings to take at least m wickets on k occasions
- Innings to take at least n wickets from debut
- Innings by innings bowling list for a selected player

Below are shown a few sample charts. Clicking on the chart will take you to a new page where you can interact further.

**Test batting:**

**Test bowling:**

**ODI batting:**

**ODI bowling:**

It can be readily seen that one can obtain a parallelogram having half the area of a triangle by connecting a vertex with the three midpoints of the sides. (This has half the base length of the triangle and half its height.)

Is it possible to obtain a larger parallelogram? As outlined in [1], if two or fewer vertices of the parallelogram are on sides of the triangle, a smaller similar triangle can be created by drawing a line parallel to the triangle’s side through a vertex of the parallelogram that is interior triangle. (This is done three times in the figure below.) This reduces the problem to the next case.

We are left to consider the case where three or more vertices of the parallelogram on the triangle. We can draw a line from a vertex to the opposite side parallel to a pair of sides (in the figure below is drawn parallel to ), thus dissecting the triangle into two. Each of the smaller triangles then has an inscribed parallelogram where two of the vertices are on a side of each triangle. Then by drawing lines parallel to sides if required, we create two sub-problems each having four vertices of the parallelogram on the sides of the triangle.

Finally, if all four vertices of the parallelogram are on sides of the triangle, by the pigeonhole principle, two of them are on a side (say as shown in the figure below). In this figure, if we let and the height of ABC from be , then by the similarity of triangles and , and has height . Then the area of the parallelogram is which is times the area of . This quantity has maximum value 1/2 when so we conclude that the parallelogram does not exceed half the triangle’s area.

Constraining sides of the parallelogram to be equal (forming a rhombus), we claim that the largest rhombus that can be inscribed in a triangle is also half its area. This can be formed with two of the vertices on the second longest side of the triangle. Suppose are the sides of the triangle with the second longest side length. Then let the segment joining the midpoints of and form one side of the rhombus of length . It remains to be shown that there exist parallel segments of this same length from to . The longest possible such segment has length and the shortest has length , half the length of the altitude of from . We wish to show that . This follows from

The area of this rhombus is clearly half the area of the triangle as it has half the length of its base and half the height.

Here if is the longest side of the triangle we form the rectangle from midpoints of and respectively, dropping perpendiculars onto forming rectangle :

The area of this rectangle is half the area of the triangle as it has half the length of its base and half the height.

Interestingly the reflections of the vertices of the triangle in the sides of the rectangle coincide, showing a paper folding interpretation of this result [2].

Since rhombuses and rectangles are special cases of parallelograms and we found that inscribed parallelograms in a triangle occupy no more than half its area, the rhombus and rectangle constructions here are optimal.

Here we shall see that the best we can do may not be half the area of the triangle. As before, if two or fewer vertices of the square are not on the sides of the triangle it is possible to scale up the square (or scale down the triangle) so that three of the square’s vertices are on the sides. We claim that the largest square must have two of its vertices on a side of the triangle. Suppose this is not the case and we have the figure below.

Consider squares inscribed in so that one vertex is on , is on and is on . We claim that the largest such square is either (two vertices on ) or (two vertices on ). Suppose on the contrary that neither of these squares is the largest. Then we make use of the fact that all 90-45-45 triangles inscribed in have a common pivot point . This is the point at the intersection of the circumcircles of triangles , and . To show these circles intersect at a single point, we can prove that if the circumcircles of triangles and intersect at then the points are cyclic by the following equality:

where the second last equality makes use of quadrilaterals , being cyclic.

Additionally we have

with similar relations for and . Hence is the unique point satisfying

(Each equation defines a circular arc, they intersect at a single point. Note that may be outside triangle .) This point is the centre of spiral similarity of 90-45-45 triangles with respectively on the sides of the triangle. Consider the locus of the points of the square as vary on straight line segments pivoting about . It follows that the fourth point of the square also traces a line segment, between the points and so as to be contained within the triangle.

As the side length of the square is proportional to the distance of a vertex to its pivot point, the largest square will be where is maximised. We have seen that the point varies along a line segment, so will be maximised at one of the extreme points – either when or . We therefore conclude that the largest square inside a triangle will have two points on a side.

If the triangle is acute-angled, by calculating double the area of the triangle in two ways, the side length of a square on the side of length with altitude is derived as

If the triangle is obtuse-angled, the square erected on a side may not touch both of the other two sides. In the figure below the side length of square is the same as if were moved to , where is right-angled. In this case the square’s side length is .

The largest square erected on a side may be constructed using the following beautiful construction [2]: simply erect a square CBDE external to the side and find the intersection points .

These points define the base of the square to be inscribed since by similar triangles

so that

One can use this interactive demo to view the largest square in any given triangle. One needs to find the largest of the three possibilities of the largest square erected on each side. In the acute-angled-triangle case, the largest square is on the side that minimises the sum of that side length and its corresponding perpendicular height – as their product is fixed as twice the triangle’s area, this will occur when the side and height have minimal difference. For a right-angled triangle with legs and hypotenuse , we wish to compare the quantities and , the two possible sums of the base and height of the triangle. We always have because the diameter of the incircle of the triangle is shorter than the altitude from the hypotenuse (i.e. the incircle is inside the triangle). We conclude that the largest square in a right-angled triangle is constructed on its two legs rather than its hypotenuse.

[1] I. Niven, *Maxima and Minima without Calculus*, The Mathematical Association of America, 1981.

[2] M. Gardner, *Some Surprising Theorems About Rectangles in Triangles*, Math Horizons, Vol. 5, No. 1 (September 1997), pp. 18-22.

[3] Jaime Rangel-Mondragon *“Largest Square inside a Triangle”* http://demonstrations.wolfram.com/LargestSquareInsideATriangle/ Wolfram Demonstrations Project Published: March 7 2011

(Click on the above image to go to the Tableau page if you wish to change the parameters. You can also select the “Innings by innings” tab to look up a player’s list of innings.)

Below we some examples for different m (full-career stats among players who have played at least 20 innings).

**m=1**: A total of 22 players have had an entire career of 20+ innings getting off the mark each time, with RA Duff (Australia, 1902-1905) the only to play 40 innings (note that JW Burke played 44 innings without a duck, but made 0 not out in one innings).

Angelo Mathews (SL) has managed just 2 ducks in his 154 innings to date.

**m=10**: Hobbs, Hutton, Kanhai and Sobers stand out here, having played over 100 innings and reaching double figures at least 80% of the time (Hobbs over 86%). Labuschagne, Hetmyer, Handscomb and Head are recent players to feature highly here.

**m=25**: Bradman starts to distance himself from the rest. Hammond, Smith, Sobers and de Villiers also impress here.

**m=50**: Smith has matched Barrington’s career figures of 50+ starts. Sutcliffe had 33 50+ scores in his first 64 innings, the same as Bradman.

**m=75**: Barrington’s numbers are amazing here and Katich is ahead of Kohli, Tendulkar and Lara.

**m=100**: Smith and Kohli are currently higher than Sangakkara, the highest among recent retirees.

**m=125**: Only Bradman (6) had more 140+ scores than Labuschagne after playing 23 test innings, equal with Graeme Smith (who had 4 150+ scores in his first 17 innings!).

**m=150**: Bradman is so far ahead of the rest here. Lara and Sangakkara are just one behind Tendulkar with the most 150+ scores despite almost 100 fewer innings.

**m=175**: Again Lara and Sangakkara have the same number of 175+ scores (15).

**m=200**: Kohli is similar to Hammond’s career figures at this stage, with 6 of his 7 double centuries coming within 33 innings between July 2016 and December 2017.

Please leave any other interesting observations in the comments.

Line AC:

Incentre:

Centroid:

Circumcentre:

Orthocentre:

Nine-point centre: (midpoint of the midpoints of AB and BC)

Angle bisectors:

Ex-centres (intersection of internal and external bisectors):

Lines joining the excentres (in red above):

Altitude to the hypotenuse:

Euler line:

Foot of altitude to the hypotenuse: (where intersects )

Symmedian point (midpoint of the altitude to the hypotenuse [1]):

Contact points of incircle and triangle:

Gergonne point (intersection of Cevians that pass through the contact points of the incircle and triangle = the intersection of and ):

Nagel point (intersection of Cevians that pass through the contact points of the ex-circles and triangle = the intersection of and :

[1] Weisstein, Eric W. “Symmedian Point.” From *MathWorld*–A Wolfram Web Resource. http://mathworld.wolfram.com/SymmedianPoint.html

Temperature (°C) | Date | Station Name | State |

50.7 | 2-Jan-60 | Oodnadatta Airport | SA |

50.5 | 19-Feb-98 | Mardie | WA |

50.3 | 3-Jan-60 | Oodnadatta Airport | SA |

49.9 | 19-Dec-19 | Nullarbor | SA |

49.8 | 19-Dec-19 | Eucla | WA |

49.8 | 21-Feb-98 | Emu Creek Station | WA |

49.8 | 13-Jan-79 | Forrest Aero | WA |

49.8 | 3-Jan-79 | Mundrabilla Station | WA |

49.7 | 10-Jan-39 | Menindee Post Office | NSW |

49.6 | 12-Jan-13 | Moomba Airport | SA |

49.5 | 19-Dec-19 | Forrest | WA |

49.5 | 24-Jan-19 | Port Augusta Aero | SA |

49.5 | 24-Dec-72 | Birdsville Police Station | QLD |

49.4 | 21-Dec-11 | Roebourne | WA |

49.4 | 16-Feb-98 | Emu Creek Station | WA |

49.4 | 7-Jan-71 | Madura Station | WA |

49.4 | 2-Jan-60 | Marree Comparison | SA |

49.4 | 2-Jan-60 | Whyalla (Norrie) | SA |

49.3 | 27-Dec-18 | Marble Bar | WA |

49.3 | 2-Jan-14 | Moomba Airport | SA |

49.3 | 9-Jan-39 | Kyancutta | SA |

49.2 | 20-Dec-19 | Keith West | SA |

49.2 | 24-Jan-19 | Kyancutta | SA |

49.2 | 21-Feb-15 | Roebourne Aero | WA |

49.2 | 10-Jan-14 | Emu Creek Station | WA |

49.2 | 22-Dec-11 | Onslow Airport | WA |

49.2 | 1-Jan-10 | Onslow | WA |

49.2 | 11-Jan-08 | Onslow | WA |

49.2 | 9-Feb-77 | Mardie | WA |

49.2 | 1-Jan-60 | Oodnadatta Airport | SA |

49.2 | 3-Jan-22 | Marble Bar Comparison | WA |

49.2 | 11-Jan-05 | Marble Bar Comparison | WA |

49.1 | 24-Jan-19 | Tarcoola Aero | SA |

49.1 | 23-Jan-19 | Red Rocks Point | WA |

49.1 | 13-Jan-19 | Marble Bar | WA |

49.1 | 27-Dec-18 | Onslow Airport | WA |

49.1 | 3-Jan-14 | Walgett Airport AWS | NSW |

49.1 | 2-Jan-10 | Emu Creek Station | WA |

49.1 | 18-Feb-98 | Roebourne | WA |

49.1 | 23-Dec-72 | Moomba | SA |

49 | 15-Jan-19 | Tarcoola Aero | SA |

49 | 23-Jan-15 | Marble Bar | WA |

49 | 13-Jan-13 | Birdsville Airport | QLD |

49 | 9-Jan-13 | Leonora | WA |

49 | 21-Dec-11 | Roebourne Aero | WA |

49 | 1-Jan-10 | Mardie | WA |

49 | 10-Jan-09 | Emu Creek Station | WA |

49 | 11-Jan-08 | Port Hedland Airport | WA |

49 | 11-Jan-08 | Roebourne | WA |

49 | 12-Jan-88 | Marla Police Station | SA |

49 | 6-Dec-81 | Birdsville Police Station | QLD |

49 | 22-Dec-72 | Marree | SA |

[1] http://www.bom.gov.au/cgi-bin/climate/extremes/monthly_extremes.cgi

[2] http://www.bom.gov.au/climate/current/statements/scs43e.pdf

]]>Kinetic energy is that associated with motion and is defined as for a particle with mass , velocity and momentum . If the mass is a fluid in motion (e.g. wind) with density and volume through cross-sectional area , then .

Work is the result of a force applied over a displacement and is given by the line integral

This has the simple form when force is constant and displacement is linear where is the angle between the force and displacement vectors.

Using Newton’s 2nd law and the relation this can be written as

This is the work-energy theorem which says that work is the change in kinetic energy by a net force. It can also be written as where is momentum.

The above has the rotational analogue where is moment of inertia and is angular velocity and the equation for work becomes

,

where is a torque vector.

This has the simple form in the special case of a constant magnitude tangential force where is the torque resulting from force applied at distance from the centre of rotation.

Note that the time derivative of work is defined as power, so work can also be expressed as the time integral of power:

If the work done by a force field depends only on a particle’s end points and not on its trajectory (i.e. conservative forces), one may define a potential function of position, known as potential energy satisfying . By convention positive work is a reduction in potential, hence the minus sign. It then follows that in such force fields the sum of kinetic and potential energy is conserved.

Some types of potential energy:

- due to a gravitational field: , where are the masses of two bodies, the distance between their centre of masses and is Newton’s gravitation constant.
- due to earth’s gravity at the surface: where and is the object’s height above ground (small compared with the size of the earth).
- due to a spring obeying Hooke’s law: where is the spring constant and the displacement from an equilibrium position.
- due to an electrostatic field: where is Coulomb’s constant and are charges. This can be written as where is a potential function measured in volts.
- for a system of point charges: .
- for a system of conductors: where the charge on conductor is and its potential is .
- for a charged dielectric: the above may be generalised to the volume integral where is charge density and is the potential corresponding to the electric field.
- for an electric dipole in an electric field: where is directed from the negative to positive charge and has magnitude equal to the product of the positive charge and charge separation distance.
- for a current loop in a magnetic field: where is directed normal to the loop and has magnitude equal to the product of the current through the loop and its area.

In electric circuits the voltage drop across an inductance is and the current though a capacitance is . These inserted into the relationship lead to the formulas and for the energy stored in a capacitor and inductor respectively.

Also in electromagnetism the energy flux (flow per unit area per unit time) is the Poynting vector , the cross product of the electric and magnetising field vectors. The electromagnetic energy in a volume is given by ([1])

,

where is the electric displacement field and is the magnetic field. This is more commonly written as when the relationships hold.

In special relativity energy is the time component of the momentum 4-vector. That is, energy and momentum are mixed in a similar way to how space and time are mixed at high velocities. Computing the norm of the momentum four-vector gives the energy-momentum relation

.

This leads to for massless particles (such as photons) and more generally , the mass-energy equivalence relation (here and is rest mass).

In quantum mechanics the energy of a photon is also written as (Planck-Einstein relation) where is Planck’s constant and are frequency and wavelength respectively. Energies of quantum systems are based on the eigenstates of the Hamiltonian operator, an example of which is .

Force is also equal to pressure times area, so another formula for work (e.g. done by an expanding gas) is the volume integral . In thermodynamics heat is energy transferred through the random motion of particles. The fundamental equation of thermodynamics quantifies the internal energy which disregards kinetic or potential energy of a system as a whole (only considering microscopic kinetic and potential energy):

where is temperature, is entropy, is the number of particles and the chemical potential of species . Similar formulas exist for other thermodynamic potentials such as Gibbs energy, enthalpy and Helmholtz energy.

The mean translational kinetic energy of a bulk substance is related to its temperature by where is Boltzmann’s constant.

In thermal transfer the change in internal energy is given by where is mass and is the heat capacity which may apply to constant volume or constant pressure.

The power per unit area emitted by a body is given by the Stefan-Boltzmann law where is the emissivity (=1 for black body radiation) and is the Stefan–Boltzmann constant. This equation may be used to determine the energy emitted by stars using their emission spectrum.

The latent heat (thermal energy change during a phase transition) of mass of a substance with specific latent heat constant is given by .

Finally, the energy of a single wavelength of a mechanical wave is where the mass of a wavelength, the amplitude and the angular frequency [2]. This can be applied to finding the energy density of ocean waves for example [3].

[1] Poynting Vector. *Brilliant.org*. Retrieved 22:24, December 28, 2018, from https://brilliant.org/wiki/poynting-vector/

[2] Power of a Wave. *Brilliant.org*. Retrieved 21:23, December 30, 2018, from https://brilliant.org/wiki/power-of-a-wave/

[3] Wikipedia contributors, “Wave power,” *Wikipedia, The Free Encyclopedia,* https://en.wikipedia.org/w/index.php?title=Wave_power&oldid=875183814 (accessed December 30, 2018).

[4] Wikipedia contributors, “Work (physics),” *Wikipedia, The Free Encyclopedia,* https://en.wikipedia.org/w/index.php?title=Work_(physics)&oldid=874162163 (accessed December 30, 2018).

[5] Wikipedia contributors, “Potential energy,” *Wikipedia, The Free Encyclopedia,* https://en.wikipedia.org/w/index.php?title=Potential_energy&oldid=873393028 (accessed December 30, 2018).

[6] Wikipedia contributors, “Electric potential energy,” *Wikipedia, The Free Encyclopedia,* https://en.wikipedia.org/w/index.php?title=Electric_potential_energy&oldid=868852409 (accessed December 30, 2018).

[7] Wikipedia contributors, “Thermodynamic equations,” *Wikipedia, The Free Encyclopedia,* https://en.wikipedia.org/w/index.php?title=Thermodynamic_equations&oldid=865388931 (accessed December 30, 2018).

[8] H. Ohanian, *Physics*, 2nd edition, Norton & Company, 1989.

Win-loss breakdown by tournament (Masters tournament finals changed to best of 3 sets from 2007):

- French Open: 86-2
- Davis Cup: 18-0
- Barcelona Open: 2-0
- Monte Carlo Masters: 2-0
- Rome Masters: 2-0
- Stuttgart: 1-0

Win-loss breakdown by number of sets (overall he has won 331 and lost 36 completed sets so even winning a set against him is a big deal!):

- 5 sets: 4-0 (Coria, Federer, Isner, Djokovic – 5th set scores 7-6 (6), 7-6 (5), 6-4, 9-7 respectively)
- 4 sets: 22-1 (loss to Söderling)
- 3 sets: 83-1

Most common opponents (2 or more matches):

- Djokovic: 7-1 (lost 7 sets)
- Federer: 7-0 (lost 7 sets)
- Ferrer: 4-0 (lost 1 set)
- Almagro: 4-0
- Hewitt: 4-0 (lost 1 set)
- Söderling: 3-1 (lost 3 sets)
- Thiem: 3-0
- del Potro: 3-0 (lost 1 set)
- Gasquet: 3-0
- Seppi: 2-0 (lost 1 set)
- Murray: 2-0
- Roddick: 2-0 (lost 1 set)
- Coria: 2-0 (lost 3 sets)
- Ljubicic: 2-0
- Monaco: 2-0
- Bolelli: 2-0
- Wawrinka: 2-0 (same score of 6-2 6-3 6-1 both times)
- Bellucci: 2-0

Breakdown by set score (almost the same likelihood of winning a set 6-2, 6-3 or 6-4):

- 6-0: 26
- 6-1: 61
- 6-2: 68
- 6-3: 66
- 6-4: 67
- 7-5: 18
- 7-6: 24
- 9-7: 1
- 6-7: 9
- 5-7: 6
- 4-6: 8
- 3-6: 6
- 2-6: 3 (Federer in 2006 Rome, Söderling in 2009 French Open, Djokovic in 2012 French Open)
- 1-6: 3 (Federer in 2006 French Open, del Potro in 2011 Davis Cup, Djokovic in 2015 French Open)
- 0-6: 1 (Coria in 2005 Monte Carlo Masters)

(only one incomplete set 2-0 after which Pablo Carreno Busta retired)

(1) Tennis Abstract: Rafael Nadal ATP Match Results, Splits, and Analysis

]]>**Example 1**: The number of ways of colouring an m by n grid black or white so that there is an even number of 1s in each row and column is

**Proof**: The first rows and columns may be coloured arbitarily. This then uniquely determines how the bottom row and rightmost column are coloured (restoring even parity). The bottom right square will be black if and only if the number of black squares in the remainder of the grid is odd, hence this is also uniquely determined by the first rows and columns. Details are also given here.

**Example 2**: The number of ways of colouring an m by n grid black or white so that every 2 by 2 square has an odd number (1 or 3) of black squares is

**Proof**: First colour the first row and first column arbitarily (there are such squares each with 2 possibilities). This uniquely determines how the rest of the grid must be coloured by considering the colouring of adjacent squares above and to the left.

By the same argument, the above is the same as the number of colouring an m by n grid black or white so that every 2 by 2 square has an even number (0, 2 or 4) of black squares.

**Example 3**: The number of ways of colouring an m by n grid black or white so that every 2 by 2 square has two of each type is

**Proof**: If there are two adjacent squares of the same colour with one above the other, the remaining squares of the corresponding two rows are uniquely determined as being the same alternating between black and white. The remainder of the grid is then determined by the colouring of first column ( possibilities where we omit the two cases of alternating colours down the first column). Such a grid cannot have two horizontally adjacent squares of the same colour. By a similar argument a colouring that has two adjacent colours with one left of the other can be done in ways. Finally we have the two additional configurations where there are no adjacent squares of the same colour, which is uniquely determined by the colour of the top left square. Hence in total we have possible colourings.

This question for was in the 2017 Australian Mathematics Competition and the general solution is also discussed here.

**Example 4**: The number of ways of colouring an m by n grid black or white so that each row and each column contain at least one black square is (OEIS A183109)

Proof: First we count the number of colourings where a fixed subset of columns is entirely white and each row has at least one black square. The remaining columns and rows can be coloured in ways. To count colourings where each column has at least one black square we apply the principle of inclusion-exclusion and arrive at the above result.

Another inclusion-exclusion example shown here counts the number of 3 by 3 black/white grids in which there is no 2 by 2 black square. The answer is 417 with more terms for n by n grids in OEIS A139810.

**Example 5**: Suppose we wish to count the number of colourings of an m by n grid in which row i has black squares and column j has black squares (, ). Following [1], the number of ways this can be done is the coefficient of in the polynomial

To see this note that expanding the product gives products of terms of the form where such a term included corresponds to the ‘th row and th column being coloured black. Hence the coefficient of is the number of ways in which the system has a solution (, ) for equal to 1 if and only if row and column are coloured black and 0 otherwise.

Let us evaluate this in the special case of 2 black squares in every row and every column for an n by n square grid (i.e. and ). Picking two squares in each column to colour black means viewing the expansion as a polynomial in the coefficient of has sums of products of terms of the form . Then using notation to denote the coefficient of an expression, we have

Here the second last line follows from considering the number of ways that products of terms of the form arise in the product (which is ) and products of terms of the form can be formed in the product (which is ).

For example, when this is equivalent to finding the coefficient of in . Products are either paired up in complementary ways such as in ( ways) or we have the three products ( ways). This gives us a total of 90 (this question appeared in the 1992 Australian Mathematics Competition). More terms of the sequence are found in OEIS A001499 and the 6 by 4 case (colouring two shaded squares in each row and three in each column in 1860 ways) appeared in the 2007 AIME I (see Solution 7 here).

**Example 6**: If we wish to count the number of grid configurations in which reflections or rotations are considered equivalent, we may make use of Burnside’s lemma that the number of orbits of a group is the average number of points fixed by an element of the group. For example, to find the number of configurations of 2 by 2 grids up to rotational symmetry, we consider the cyclic group . For quarter turns there are configurations fixed (a quadrant determines the colouring of the remainder of the grid) while for half turns there are configurations as one half determines the colouring of the other half. This gives us an answer of

which is part of OEIS A047937. If reflections are also considered equivalent we need to consider the dihedral group and we arrive at the sequence in OEIS A054247.

If we want to count the number of 3 by 3 grids with four black squares up to equivalence, this is equivalent to the number of full noughts and crosses configurations. A nice video by James Grime explaining this is here (the answer is 23).

**Example 7**: The number of ways of colouring an m by n grid black or white so that the regions form 2 by 1 dominoes has the amazing form

For example, the 36 ways of tiling a 4 by 4 grid are given here. A proof of the above formula using the Pfaffian of the adjacency matrix of the corresponding grid graph is given in chapter 10 of [2].

[1] L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions (pp 235-6), D. Reidel Publishing Company, 1974.

[2] M. Aigner, A Course in Enumeration, Springer, 2007.

]]>Note that the first data point in 1974 corresponds to a game that was reduced to 50 overs per side after originally intended to be a 55 over game.

If we slice the data into eras marked by calendar years of roughly equal numbers of games, the mean score had a slight slow-down in the rate of increase from 2008-2012, then accelerated again in the past five years.

Era | Number of matches | Mean score batting first |

1974-1994 | 427 | 229 |

1995-1999 | 383 | 247 |

2000-2003 | 368 | 257 |

2004-2007 | 380 | 267 |

2008-2012 | 393 | 272 |

2013-2017 | 398 | 288 |

1974-2017 |
2349 |
260 |

The histograms below show how rarely teams score less than 200 runs in recent times when using the full quota of 50 overs. In fact these days a team is more likely to score over 400 than below 200 if using the full quota of 50 overs!

Comparing the distribution of first innings winning versus losing scores we find that the mean scores are 275 vs 236 respectively with sample sizes 1392 vs 901 (34 games had no result and 22 were tied). Restricting to the past five years, the median score batting first for the full 50 overs in winning matches is exactly 300.

Interestingly if we break down the runs scatter plot by team, the trends are not the same across the board. In particular England and South Africa have had more dramatic increases in recent times than the other teams, especially compared with India, Pakistan, Sri Lanka and West Indies.

Restricting to the last five years (2013-2017), here are the mean first innings scores for each team based on the match result (assuming they bat the full 50 overs).

Team | Result | mean score | # matches |

Afghanistan | lost | 249 | 6 |

Afghanistan | won | 260 | 12 |

Australia | lost | 295 | 13 |

Australia | n/r | 253 | 3 |

Australia | won | 310 | 31 |

Bangladesh | lost | 263 | 16 |

Bangladesh | won | 275 | 15 |

Canada | lost | 230 | 3 |

England | lost | 282 | 13 |

England | won | 329 | 22 |

Hong Kong | won | 283 | 4 |

India | lost | 282 | 12 |

India | won | 310 | 27 |

Ireland | lost | 244 | 6 |

Ireland | tied | 268 | 1 |

Ireland | won | 289 | 3 |

Kenya | lost | 260 | 1 |

Netherlands | lost | 265 | 1 |

New Zealand | lost | 277 | 12 |

New Zealand | tied | 314 | 1 |

New Zealand | won | 308 | 27 |

P.N.G. | lost | 218 | 2 |

P.N.G. | won | 232 | 1 |

Pakistan | lost | 266 | 9 |

Pakistan | n/r | 296 | 1 |

Pakistan | tied | 229 | 1 |

Pakistan | won | 290 | 20 |

Scotland | lost | 238 | 6 |

Scotland | won | 284 | 8 |

South Africa | lost | 258 | 7 |

South Africa | n/r | 301 | 1 |

South Africa | won | 321 | 36 |

Sri Lanka | lost | 249 | 15 |

Sri Lanka | n/r | 268 | 2 |

Sri Lanka | tied | 286 | 1 |

Sri Lanka | won | 305 | 22 |

U.A.E. | lost | 279 | 3 |

U.A.E. | won | 267 | 3 |

West Indies | lost | 265 | 10 |

West Indies | won | 298 | 10 |

Zimbabwe | lost | 247 | 9 |

Zimbabwe | tied | 257 | 1 |

Zimbabwe | won | 276 | 1 |

The England and South Africa numbers stand out the most here in winning causes. Also Australia has a particularly high average score of 294 in losing causes. Sri Lanka has the largest difference (56 runs) between average winning and losing scores.

**Edit**: The following shows the mean scores in the 100 matches prior to and after key rule changes (still focusing on first innings 50-over scores). Note that in two of the three cases, the average scores reduced.

- Restriction of 2 outside the 30-yard circle in the first 15 overs (’92 World Cup)

03 Jan 88 to 20 Jan 92:**231**

12 Feb 92 to 16 Feb 94:**222** - Introduction of Powerplay overs

13 Mar 04 to 30 Jun 05:**267**

07 Jul 05 to 08 Sep 06:**267** - Removal of powerplay, fifth fielder allowed outside the circle in the last ten overs

17Aug 14 to 24 Jun 15:**301**

10 Jul 15 to 19 Jan 17:**289**