Mathematics, Everywhere

By Chandrahas M. Halai

लौकिके वैदिके वापि तथा सामायिकेऽपि य: ।

व्यापारस्तत्र सर्वत्र सङ्ख्यानमुपयुज्यते ॥ ९

कामतन्त्रेऽर्थशस्त्रे च गान्धर्वे नाटकेऽपि वा।

सूपशास्त्रे तथा वैद्ये वास्तुविद्यादिवस्तुषु ॥ १०

छन्दोऽलङ्कारकाव्येषु तर्कव्याकरणादिषु ।

कलागुणेषु सर्वेषु प्रस्तुतं गणितं परम् ॥ ११

सूर्यादिग्रहचारेषु ग्रहणे ग्रहसंयुतौ ।

त्रिप्रश्ने चन्द्रवृतौ च सर्वत्राङ्गीकृतं हि तत् ॥ १२

In worldly life, in Vedic learning, in religious practice, in business, in everything, Mathematics is useful. In romance, economics, in music, dance and drama, in cooking, medicine and in architecture, in prosody, poetry, logic and grammar, and in all the arts, Mathematics reigns supreme. It is used in calculating movements of the Sun and other heavenly bodies.

बहुभिर्विप्रलापै: किं त्रिलोक्ये सचराचरे ।

यत्किञ्चिद्वस्तु तत्सर्वं गणितेन विना न हि ॥ १६

Whatever there is in all the three worlds, which are possessed by moving and non-moving beings- all that indeed cannot exist as apart from Mathematics.

The above verses are from the book Ganitasarasangraha written around 850 CE by Digambara Jain monk Mahavira (815 – 877 CE).

You might be wondering what has art got to do with Mathematics? And of all the things what has mathematics got to do with cooking?

As an answer to this, lo behold, here is a problem from Bhaskaracharya’s (Bhaskaracharya II) (Born 1114 CE) book Lilavati :

एकद्वित्र्यादियुक्त्या मधुरकटुकषायाम्लकक्षारतिक्तैः

एकस्मिन् षड्रसैः स्युर्गणक कति वद व्यंजने व्यक्तिभेदाः।।

In how many different ways can we garnish a dish using six different types of flavouring substances – sweet, bitter, salty, sour, pungent and hot, taking them by ones, twos, threes, and so on at a time?

Bhaskaracharya had composed Lilavati in 1150 CE to teach Mathematics to his daughter Lilavati, after whom he had named the book. This book which is written entirely in verse covers arithmetic, algebra, geometry, mensuration, combinatorics, number theory and other basic topics. This book was used as a standard textbook for almost 700 years.

Coming back to the above problem –

I may not garnish my dish at all. Now, this can only be done in one way. Remember this subtotal, 1.

I can garnish the dish using all the six ingredients. Now, this can be done in only one way. Remember this subtotal, 1.

I may garnish the dish using any one of the six flavouring substances. This can be done in six different ways. Subtotal = 6.

I may garnish the dish using any five of the six ingredients. This can be achieved by leaving out anyone of the six ingredients. Now, this can be done in six different ways. Subtotal = 6.

Now, I want to garnish the dish using any two of the six flavouring substances. In how many different ways can this be done?

In the same chapter of Lilavati from which the given problem is taken, Bhaskaracharya has given us the formula for calculating the number of combinations of ‘n‘ things taken ‘r‘ at a time as

Using modern notation the same formula is given as

Using this formula let’s calculate in how many different ways I can garnish my dish using any two of the six different flavouring substances

Hence, this can be done in 15 different ways. Subtotal = 15.

I may garnish my dish using any three of the six ingredients. This can be done in

different ways. Hence, subtotal = 20.

I may garnish my dish using any four of the six ingredients. This can be done in

different ways. Hence, subtotal = 15.

Hence, I can garnish my dish in = 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64 different ways.

This was the kind of awareness amongst ancient and medieval Indians about mathematics and its applications in daily life.

Ancient Indians considered Mathematics as the most significant subject. This is evident from this verse from Vedanga Jyotisa (500 BCE).

यथा शिखा मयूराणां , नागानां मणयो यथा ।

तद् वेदांगशास्त्राणां , गणितं मूर्ध्नि वर्तते ॥

Like the crest of a peacock, like the gem on the head of a snake,

so is mathematics at the head of all knowledge.

An Oily Matter, part 2

By Chandrahas M. Halai

This is in continuation of my earlier article:

https://chandrahasblogs.wordpress.com/2020/10/21/its-an-oily-matter/

How many measures of oil would the devotee have to bring to the temple if there was a row of sixteen oil lamps?

Solution:

Carrying forward our calculations:

At the Sixth Oil Lamp:

The devotee brings 32x -124 measures of oil to the sixth lamp.

The attendant at the sixth oil lamp adds equal amount of oil, 32x-124 measures to the devotee’s can.

Therefore, now the amount of oil in the devotee’s can = 2(32x-124) = 64x – 248 measures.

The devotee pours 4 measures of oil into the fifth lamp. Therefore, now the oil remaining in the can = (64x – 248) – 4 = 64x – 252 measures.

Now wait! Are we going to carry on this process till we empty the oil in the devotee’s can at the sixteenth oil lamp? That would be quite a tedious process. What if there was a row of thirty-six oil lamps? Forget it; I don’t want to solve this problem.

Is there a pattern in this process? Let us try and find it.

After pouring Oil in the First Lamp:

The oil remaining in the can

After pouring Oil in the Second Lamp:

The oil remaining in the can

After pouring Oil in the Third Lamp:

The oil remaining in the can

After pouring Oil in the Fourth Lamp:

The oil remaining in the can

So now, don’t we see a definite pattern here?

Therefore, the amount of oil remaining in the can after pouring it in the n^th lamp

Suppose the n^th lamp is the last one, then the devotee should have emptied his can at this lamp. Hence, we have

This is the amount of oil that the devotee has to take to the temple if there are n oil lamps in the temple.

Therefore, if there are 16 oil lamps in the temple, then the devotee has to take carry

Now let us analyze the formula we got

Here, if n becomes larger, gets closer and closer to 1. This means that the amount of oil the devotee takes to the temple gets closer to 4 measures.

What if there are infinite numbers of oil lamps in the temple?

Then the amount of oil that the devotee has to take to the temple is 4 measures.

Let me put it differently.

The devotee takes four measures of oil to the temple. At the first oil lamp the attendant adds another 4 measures of oil into the devotee’s can. Now, there are 8 measures of oil in the can. The devotee pours 4 measures of oil in the first lamp. Hence, now there are of 4 measures of oil remaining in the can, which he carries to the second oil lamp. Here too, the attendant adds 4 measures of oil to the devotee’s can. Now the can contains 8 measures of oil. Out of this the devotee pours 4 measures of oil into the second lamp. Therefore, 4 measures of oil remain in the can, which the devotee carries to the third lamp. This continues ad infinitum.

It’s an Oily Matter

By Chandrahas M. Halai

A temple in a town in India has a row of five large oil lamps which leads the devotee from the entrance to the deity in the sanctum sanctorum. Each oil lamp can take four measures of oil. This temple has a unique practice. When a devotee arrives at the temple with some oil in a can, an attendant at the first lamp from the entrance pours an equal amount of oil into the can. So, the can now contains double the amount of oil the devotee brought. The devotee then pours 4 measure of oil into the first lamp and moves on with the remaining oil to the second lamp.

The attendant at the second lamp, too, gives him the same amount of oil he has in the can. So the devotee now has double the amount of oil he brought to the second lamp. He pours 4 measures of oil into the second lamp and moves on with remaining oil to the third lamp.

At the third and fourth lamp the same procedure is repeated.

At the fifth lamp which is the last lamp, he receives an equal amount of oil from the attendant present there. Now the devotee has exactly four measures of oil in his can. He pours all of it into the fifth lamp and reaches the deity with an empty can.

How many measures of oil did the devotee bring to the temple?

Solution:

Let the amount of oil in the can the devotee brought to the temple be x measures.

At the First Oil Lamp:

The attendant at the first oil lamp adds equal amount of oil, x measures to the devotee’s can.

Therefore, now the amount of oil in the devotee’s can = x + x = 2x measures.

The devotee pours 4 measures of oil into the first lamp. Therefore, now the oil remaining in the can = 2x – 4 measures.

At the Second Oil Lamp:

The devotee brings 2x – 4 measures of oil to the second lamp.

The attendant at the second oil lamp adds equal amount of oil, 2x – 4 measures to the devotee’s can.

Therefore, now the amount of oil in the devotee’s can = 2(2x – 4) = 4x – 8 measures.

The devotee pours 4 measures of oil into the second lamp. Therefore, now the oil remaining in the can = (4x – 8) – 4 = 4x – 12 measures.

At the Third Oil Lamp:

The devotee brings 4x – 12 measures of oil to the third lamp.

The attendant at the third oil lamp adds equal amount of oil, 4x – 12 measures to the devotee’s can.

Therefore, now the amount of oil in the devotee’s can = 2(4x – 12) = 8x – 24 measures.

The devotee pours 4 measures of oil into the third lamp. Therefore, now the oil remaining in the can = (8x – 24) – 4 = 8x – 28 measures.

At the Fourth Oil Lamp:

The devotee brings 8x – 28 measures of oil to the fourth lamp.

The attendant at the fourth oil lamp adds equal amount of oil, 8x – 28 measures to the devotee’s can.

Therefore, now the amount of oil in the devotee’s can = 2(8x – 28) = 16x – 56 measures.

The devotee pours 4 measures of oil into the fourth lamp. Therefore, now the oil remaining in the can = (16x – 56) – 4 = 16x – 60 measures.

At the Fifth Oil Lamp:

The devotee brings 16x – 60 measures of oil to the fifth lamp.

The attendant at the fifth oil lamp adds equal amount of oil, 16x – 60 measures to the devotee’s can.

Therefore, now the amount of oil in the devotee’s can = 2(16x – 60) = 32x – 120 measures.

The devotee pours 4 measures of oil into the fifth lamp. Therefore, now the oil remaining in the can = (32x – 120) – 4 = 32x – 124 measures.

Now, we have that the devotee has emptied his can at the fifth lamp. Thus, we have

32x – 124 = 0

32x = 124

x = 31/8

Hence, we can say that the devotee had brought 31/8 measures of oil to the temple.

Why is negative times negative, a positive?

By Chandrahas M. Halai

The most common answer to this question is that this is a rule. But in mathematics every rule has a logical reason.

Let us examine or have a fresh look at what is multiplication, what are negative numbers and how multiplication between a positive and a negative numbers is defined and lastly how multiplication of a negative number with a negative number is defined.

Counting

In the very beginning, humans invented natural numbers (or is it that they discovered them, or is it both) for the purpose of counting and measurement. One can imagine a scenario where early human needed numbers to keep a tab on his flock of cattle or horses. Cave paintings by early humans show us that they used to keep a score of their kills during their various hunting expeditions by painting lines on the walls of the caves they used to dwell in or by etching notches/lines on dead animal bones ^[1].

When humans started living in groups or tribes and there was a conflict between two tribes/groups the strategist/chieftain needed to compare the number of fight worthy men in both the tribes to decide whether it is wise to fight the battle or make a truce or make good a retreat.

Addition

As the human population grew and it started living in communities and settling in villages or small towns, so did the number of live stock and other things in its possession. People started exchanging or bartering things. Let us say a tribe chieftain wants to know the total of all livestock/cattle or horses they have at all the stables in the village. Or a General wants to know the total number of guards present at all the gates or watch towers of a fortification or a chief wanting to know the population of his tribe after merger with various small groups or after a split. Simple counting of things became tedious and time consuming, this scenario warranted the development of a method/algorithm to aid counting. These led to the development of Addition as an aid or extension to counting. The development of addition also led to the development of Subtraction as an inverse operation.

Multiplication

Soon people found that many times they were adding the same/equal numbers repeated times. You can imagine a scenario that a father, Shanti Prasad wants to buy 6 apples for each of his five children. He must add 6, five times to know the total number of apples he needs to buy from the fruit vendor. Hence Multiplication was developed as an extension of addition. Division was developed as an inverse operation to multiplication. You can visualize the multiplication process in the present example as shown in figure 1.

6 apples / child x 5 children = 30 apples

Figure 1

Distributive Rule of Multiplication

It was soon found that multiplication obeys the rule of distribution. i.e.

6 apples / child x 5 children = 6 x (3 girls + 2 boys) = 18 apples for the girls + 12 apples for the boys = 30 apples

We know that 6 x 5 = 30

Figure 2

Negative Numbers

As human civilizations evolved, they started living in cities; trade and commerce grew. Arithmetic was used to keep accounts, calculate taxes, measure areas and volumes. Algebra evolved in ancient and medieval India to deal with problems involving unknown quantities. In algebra situations arose when early mathematicians got perplexed or hit a stone wall. For e.g.

X + 5 = 2

Hence, X = 2 – 5

Early mathematicians used to wonder, how can a larger quantity be subtracted from a smaller quantity? Indian mathematicians found a way to solve such equations by inventing negative numbers.

Brahmagupta and operations with Negative Numbers

Though earlier Indian mathematicians may have worked with negative numbers, it was Brahmagupta who first wrote about the rules for doing arithmetic with negative numbers in his book Brahmasphutasiddhanta (The Opening of the Universe), in 628.

He also gives arithmetical rules in terms of fortunes (positive numbers) and debts (negative numbers):

A debt subtracted from zero is a fortune.
A fortune subtracted from zero is a debt.
The product or quotient of two fortunes is one fortune.
The product or quotient of two debts is one fortune.
The product or quotient of a debt and a fortune is a debt.
The product or quotient of a fortune and a debt is a debt.

We can imagine a scenario wherein a man returns home and tells his wife that he lost 5 silver coins at the gambling den. His wife is puzzled, how can he lose 5 coins when he had only 2. The husband clarifies that not only did he lose the 2 silver coins he had but also the 3 silver coins that he had borrowed from his friend. Hence he is now in debt of 3 silver coins.

i.e. 2 – 5 = -3

Twelfth century A.D. Indian mathematician, Bhaskaracharya also used to consider negative quantities as debt (ऋण) or losses ^[2]. Till the seventeenth century A.D. some European mathematicians used to consider negative numbers as absurd and resisted to accept their existence ^[3].

In other words the man will have to earn 5 silver coins to pay off his debt and regain his earlier financial status.

-3 + 5 = -3 + (3 + 2) = -3 + 3 + 2 = 2

Also, x (debt) + 5 = 2

Hence, x = 2 – 5 = -3

Additive Inverse

When a number is subtracted from itself we get zero. i.e.

a – a = 0    (a is a positive integer)

This is the same as a + (-a) = 0, adding a negative of a number is the same as subtracting

Two numbers having the same value but different signs (positive and negative) are called Additive Inverses of each other. In the above examples a and –a are additive inverses of each other.

Multiplication between a positive and a negative

Humans also found that many times they were subtracting the same values repeatedly. Let us again take the help of Shanti Prasad and his children. Shanti Prasad has 50 copper coins and he gives away 3 copper coins to each of his 5 children. To know how many copper coins will remain with him we have to subtract 3, five times from 50. i.e.

50 – 3 – 3 – 3 – 3 – 3 = 35

We can rewrite this repeated subtraction as

50 + (- 3 – 3 – 3 – 3 – 3) = 35

50 + (5 x -3) ?= 35

To balance this equation 5 x -3 has to be -15, hence

50 – 15 = 35

Looking at multiplication between a positive and negative in a different way, we have

a – a = 0    (a is a positive integer)

Let’s multiply both sides by a positive integer b, we get

b x (a – a) ?= b x 0

b (a – a) ?= 0

Now we know that b x a = ab, so multiplying –a (which is the additive inverse of a) by b should give us additive inverse of ab, i.e. –ab, to balance the equation and obey the Distributive rule of multiplication.

Thus, we have

ab – ab = 0

This is how multiplication between a positive and a negative number is defined, such that the multiplication obeys the distribution rule of multiplication. We can now state this as a rule:

“Multiplication of a positive number with a negative number gives us a negative number.”

And now for the final assault.

Multiplication between a negative and a negative

Let us again take the help of Shanti Prasad and his family. He has with him a treasure of 100 gold coins and he instructs his wife to give away 10 gold coins to each of his 5 children. So Shanti Prasad will be left with 100 – 5(10) = 50 gold coins after the distribution. After few days he checks his treasure and finds that he has 65 gold coins left with him. He inquires with his wife and finds that she has distributed 3 less gold coins to each of her 5 children. i.e.

100 – 5(10 – 3) = 100 – 5(7) = 100 – 35 = 65

Also, we have

100 – 5(10 – 3) ?= 65

100 + (-5 x 10) + (-5 x -3) ?= 65

100 – 50 + (-5 X -3) ?= 65

50 + (-5 x -3) ?= 65

To balance this equation -5 x -3 has to be +15, hence

50 + 15 = 65

Looking at multiplication between a negative and a negative in a different way, we have

a – a = 0    (a is a positive integer)

Let’s multiply both sides by –b (b is a positive integer), we get

–b x (a – a) ?= –b x 0

–b (a – a) ?= 0

Now we know that –b x a = –ab, so multiplying –a (which is the additive inverse of a) by –b should give us additive inverse of –ab, i.e. +ab, to balance the equation and obey the Distributive rule of multiplication.

Thus, we have

–ab + ab = 0

This is how multiplication between a negative number and a negative number is defined, such that the multiplication obeys the distribution rule of multiplication. We can now state this as a rule:

“Multiplication of a negative number with a negative number gives us a positive number.”

References:

[1] Calvin C. Clawson, The Mathematical Traveler – Exploring the Grand History of Numbers, Viva Books Pvt. Ltd., India, 2008, pp. 33-35

[2] Calvin C. Clawson, The Mathematical Traveler – Exploring the Grand History of Numbers, Viva Books Pvt. Ltd., India, 2008, pp. 126

[3] Calvin C. Clawson, The Mathematical Traveler – Exploring the Grand History of Numbers, Viva Books Pvt. Ltd., India, 2008, pp. 131-132

Chautisa Yantra, Magic Square for Peace

By Chandrahas M. Halai

 

A farmer had 16 cows. The first cow gave a litre of milk daily. The second gave two litres of milk daily, the third three litres daily and so on. Now, he wanted give away four cows to each of his four sons in such a way that the daily milk production of each of his sons is the same. Can you provide a solution to the farmer?

 

The total daily milk production of the farmer is given by = (16(16 + 1)) / 2 = 136 litres. So the total daily production for each son will be 34 litres.

 

The solution for the farmer’s problem is a 4×4 magic square. Refer figure 1.

 

7	12	1	14
2	13	8	11
16	3	10	5
9	6	15	4

 

Figure 1

 

In a 4×4 magic square the sum of numbers in each row or a column is 34. Even the sum of each 2×2 sub-squares is 34. See figure 2.

 

7	12	1	14
2	13	8	11
16	3	10	5
9	6	15	4

 

Figure 2

 

It is quite a challenge to make such a 4×4 magic square. These kinds of magic squares were being made for recreation and intellectual challenges by ancient Indian mathematicians. The magic square given in this article can be found as an inscription near the entrance of the Parshvanath Jain temple at Khajuraho. This temple was built during the 10th century AD. See the picture in figure 3.

 

 

Figure 3

 

The other feature of this magic square is that even the sum of both the diagonals comes out to be 34. Refer figure 4.

 

7	12	1	14
2	13	8	11
16	3	10	5
9	6	15	4

 

Figure 4

 

Even the sum of the split diagonals comes to be 34. Refer figures 5 (a) and (b).

7	12	1	14
2	13	8	11
16	3	10	5
9	6	15	4

 

Figure 5(a)

 

7	12	1	14
2	13	8	11
16	3	10	5
9	6	15	4

 

Figure 5(b)

 

The sum of the central 2×2 square is also 34. The sum of the four corner numbers is also 34. Adding middle two numbers of the top and bottom rows also gives you the sum 34. Same way, adding middle two numbers of the first and last columns also gives you the sum of 34. See figure 6.

 

7	12	1	14
2	13	8	11
16	3	10	5
9	6	15	4

 

Figure 6

 

The sums of the side squares as shown in figures 7 (a) and (b) is also 34.

 

7	12	1	14
2	13	8	11
16	3	10	5
9	6	15	4

 

Figure 7(a)

 

7	12	1	14
2	13	8	11
16	3	10	5
9	6	15	4

 

Figure 7(b)

 

This magic square called Chautisa Yantra (chautisa is 34 in Hindi) is believed to bring peace.

 

This inscription at Khajuraho is also an archaeological evidence of the development of the numerals (number symbols) in devanagari. At that time the rest of the world did not even know how to count.

Meru Prastaar (मेरु प्रस्तारः)

By Chandrahas M. Halai

प्रस्तारे मित्र गायत्र्याः स्युः पादे व्यक्तयः कति |

एकादिगुरवश्चाशु कथ्यतां तत्पृथक् पृथक् ||

O friend, there are six letters in the fourth line of Gayatri meter. If we choose only 1 Guru, 2 G, 3 G, 4 G, 5 G or 6 G, how many meters are possible?

This is a problem from Bhaskaracharya’s Lilavati.

How many meters with r Gurus (or Laghus) are possible in a prastaar for n syllables?

Pingala had developed a pratyay (method) to answer such questions.

Pingala’s sutra given in Chandahshastra for this method is:

परे पुर्णमिति |  – (छन्दः शास्त्रम् 8.34)

This is a very cryptic code. It was only through Halayudha Bhatt’s commentaries on Chandahshastra in his work Mritasanjeevani (composed in 10th century CE) that people were able to understand this sutra. Halayudha has explained how to create a table of numbers which he called मेरु प्रस्तार.

Table 1

Put one square in the first row. Refer Table 1. Then put 2 squares below it in the second row. Below this put 3 squares in the third row. In the fourth row put 4 squares and so on. If you want the combinations for a meter having n syllables then create a table having n + 1 rows.

Write 1 in the square in the first row. Then put 1s in the squares at the ends of each row. Refer Table 2.

Table 2

Numbers in the inner squares are got by adding the numbers in the two squares just above them. Refer Table 3. Let me explain this by calculating the number for the middle square of the third row. Both the squares above this square contain 1, hence we get 1 + 1 = 2. The number in this square gives number of meters having 1 Guru in a prastaar for 2 syllables.

Now let us calculate the value for the second square of the fourth row. The number in this square gives us the number of meters having 1 Guru in a prastaar for 3 syllables. The numbers in the two squares just above this square are 1 and 2. Hence, we get 1 + 2 = 3.

Now let us calculate the value for the third square of the fourth row. The number in this square gives us the number of meters having 2 Guru in a prastaar for 3 syllables. The numbers in the two squares just above this square are 2 and 1. Hence, we get 2 + 1 = 3.

Now let us calculate the value for the second square of the fifth row. The number in this square gives us the number of meters having 1 Guru in a prastaar for 4 syllables. The numbers in the two squares just above this square are 1 and 3. Hence, we get 1 + 3 = 4.

The seventh row of this table gives us the answer to problem posed at the beginning of this article. The first square of this row gives the number of meters when there are no Gurus. That is, all the six syllables are Laghus. This can only be done in 1 way and that is, LLLLLL.

The second square of the seventh row gives us the number of meters when there is only when one G in a quarter. This can be done in 6 different ways, viz:

1) G L L L L L

2) L G L L L L

3) L L G L L L

4) L L L G L L

5) L L L L G L

6) L L L L L G

The third square of the seventh row gives us the number of meters when there are 2 Gurus in a quarter. And so on.

Table 3

When we add up the numbers in all the squares of the seventh row it gives us the sankhya for the prastaar of 6 syllables.

S₆ = 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64

If we observe Table 3 closely, we can realise that the numbers in the squares are the binomial coefficients, ⁿC_r. Hence the number of metrical forms containing r Gurus (or Laghus) in a prastaar for n syllables is given by ⁿC_r.

Halayudha establishes that the basic rule for the construction of Meru Prastaar is the recurrence relation:

ⁿC_r = ^n-1C_r-1 + ^n-1C_r

Meru Prastaar’s form is like the steps to a hill. This must be the reason why Halayudha had called it Meru Prastaar. Meru is the name of a mountain.

Meru Prastaar is, what is popularly known as Pascal’s Triangle. But Pingala’s Meru Prastaar predates Pascal by at least 1800 years.

In the end I would like to present Bhaskaracharya’s method for calculating the binomial coefficients:

एकाधेकोत्तरा अंका व्यस्ता भाज्याः क्रमस्थितैः |

परः पूर्वेण संगुण्यस्तत्परस्तेन तेन च ||

एकद्वित्र्यादिभेदाः स्युरिदं साधारणं स्मृतम् |

छन्दश्चित्त्युत्तरे छन्दस्युपयोगोऽस्य तव्दिदाम् ॥
मूषावहनभेदादौ खण्डमेरौ च शिल्पके ।
वैध्यके रसभेदिये तन्नोक्तं विस्तृतेभर्यात् ॥

The number of combinations of n things taken r at a time are

[n (n – 1) (n – 2) … (n – (r – 1))] / [1 x 2 x 3 x … x r]

This is useful in prosody to discover all possible meters, in architecture, medical sciences, Khandmeru, chemical compositions etc. Bhaskaracharya used to call Meru prastaar as Khandmeru.