75 - One Way of Looking at the Binomial Theorem


ISSUE No. 75

A warm welcome to our new subscribers.
Vedic Mathematics is becoming increasingly popular as more and more people are introduced to the beautifully unified and easy Vedic methods. The purpose of this Newsletter is to provide information about developments in education and research and books, articles, courses, talks etc., and also to bring together those working with Vedic Mathematics. If you are working with Vedic Mathematics - teaching it or doing research - please contact us and let us include you and some description of your work in the Newsletter. Perhaps you would like to submit an article for inclusion in a later issue or tell us about a course or talk you will be giving or have given. If you are learning Vedic Maths, let us know how you are getting on and what you think of this system.

This issue's article is from Steven Vogel, a mechanical engineer from the US. The method shown in this article has an application in calculating compound interest that will be shown in the next newsletter.


The binomial theorem crops up in many interesting places in modern mathematics. It is used to expand a formula of the form (a + b)^n into individual terms. This formula has applications that range from combinatorial theory, solving certain types of probability problems, physics, and finance. You may recall the famous Pascal's Triangle, which can be used to solve for the binomial coefficients in an easy way:

      1 1
    1 2 1
  1 3 3 1
1 4 6 4 1

You may also recognize the formula:

"n choose k" not only gives you the individual coefficients, it also answers the question "how many different groups of k items can I make from a pool of n items." When the exponent of (a+b)^n is small, it's relatively easy to determine the coefficients by either Pascal's Triangle or the formula above. When the exponent n is large, however, the n! terms become very large and unwieldy. Can you imagine drawing out Pascal's Triangle for (a+b)^52 to determine how many groups of 5 cards can be drawn from a deck of 52 cards, or taking the factorial of 52 without a calculator?

At the end of chapter 32 of his book Vedic Mathematics, Bharati Krsna tells us that the binomial theorem is important, that it has many uses, and that it has a Vedic form which is more comprehensive than the current one. We will never know for sure what his method was, but I'm sure it would have been cleaner and easier than taking the factorials of large numbers. Certainly if there were a way we should be able to think about it and remember it using the sutras.

When you look closely at the equation above, you realize that many of the terms in the denominator can be divided out of the numerator, greatly reducing the effort required. Take the terms of (a+b)^4, which are shown in the last line of the triangle above. One method that I have found helpful for arriving at these terms is to use the two sutras "by one less than the one before" and "by one more than the one before." If you start with a 1, the following procedure leads to all of the binomial coefficients:

1 * 4 / 1 = 4
4 * 3 / 2 = 6
6 * 2 / 3 = 4
4 * 1 / 4 = 1

You can see how the multiplier decreases by one each time, while the divisor increases by one each time. By taking the previous term and multiplying and dividing, it is possible to arrive at each term in order.

So is it possible to use this to solve the problem of the playing cards? Can you really determine the 5th term of the expansion of (a+b)^52 this way?

1) 1
2) 1 * 52 / 1 = 52
3) 52 * 51 / 2 = 26 * 51 = 1326
4) 1326 * 50 / 3 = 442 * 50 = 22100
5) 22100 * 49 / 4 = 5525 * 49 = 270725
6) 270725 * 48 / 5 = 54145 * 48 = 2598960

All of these multiplications are easy using the "vertically and crosswise" sutra, and because the divisions are by smaller numbers it makes things easier. Certainly there is a time savings over calculating the factorial of 52.

I hope you are able to apply this technique to solve problems of your own. It is my belief that if we look at Bharati Krsna's Vedic Mathematics closely enough, there are still things left to learn and explore.





I spoke yesterday at the homeschooling convention I mentioned to you. I had an audience of over 250. There were individuals as well as families present. I broke my presentation into 3 sections:

1. General discussion how I first became aware of Vedic Math and some of the experiences I've had showing it to groups
2. A PowerPoint presentation detailing the benefits of VM and a quick technique demo of squaring numbers ending in a 5
3. Question and answer period.

It could not have gone better! I whet their appetites for VM with the discussion phase. During the PowerPoint presentation, many of the attendees present were writing down everything on each slide presented. When I demonstrated the number squaring technique, that was the clincher.

When it came time for q & a, I was a little anxious that there might not be any questions. The questions started to come fast and furious. In fact, when I ended the presentation (as the next presenter was already there setting up and I didn't want to intrude on him), I had about a dozen individuals rush the lectern wanting more information. These individuals and several others followed me to the back of the room and into the lobby to getting more information on VM.

There were several math teachers there who were very interested in bring these methods into their teaching syllabus. My wife, who was sitting at a table in the back of the room became deluged with people wanting more information. I printed up 75 copies of a 1 page description of VM and how to contact me. I felt that 75 copies would be more than enough. It turned out that it was not nearly enough as many people wanted the sheet, but, there were no more available.

One woman asked my wife if she could make copies of the page and send it out to the members of her homeschooling group as she knew they would be interested. Needless to say, my wife was thrilled at the request. I had a stack of my business cards on the table and they were all taken. Finally, I told the crowd if they wanted me to contact them directly with more information, they could write their names, e-mail addresses and/or phone numbers and I would contact them. I had 40 individuals sign the sheets for more information. Every one of them left e-mails and home phone numbers.

All in all, it was a great day for me and for Vedic Math.


This index is for the book "Vedic Mathematics" by Sri Bharati Krsna Tirthaji (1884-1960), and may be of special interest to those using the book for research. It is compiled by Steven Vogel from the 1992 Edition (2009 printing).
You can see this index at: http://www.vedicmaths.org/Introduction/History/index.asp
Please note though, different editions may have different page numberings.




Rick Blum shows how the Vedic method of multiplying numbers close to a base can be developed for multiplying numbers near different multiples of different bases. Below is an excerpt showing how to find 3002 x 204. For the full article that reviews the entire topic of Base Multiplication see: http://www.box.net/shared/u6kggatii21tyk6zt8q7

CASE III: Factors with different multiples of different bases

Try 3002 x 204. What we have here are two factors, which are multiples of different bases. It doesn't get more general than this.

1st Step: Determine Excesses and limit digits by base of lower factor

3002 + 02 Write both factors as "left justified"
204 + 04

2nd Step: Multiply vertically to obtain right side solution:

3002 + 02
204 + 04
        / 08

Step 3a) I am breaking this next step into three parts for the sake of clarity. In this first step, we multiply the smaller factor by the multiple of the larger factor's base. Looking at the larger factor, 3002, we are using a working base of 3,000. This working base is equal to 3 times a base of 1,000. Therefore, in this example, we would multiply the smaller factor, 204, by this multiple of 3:

204 x 3 = 612

Step 3b) In this 2nd step, we see that our smaller factor of 204 is 2 times a base of 100. Therefore, we then multiply this multiple of "2" times the excess generated by our larger factor we have:

2 times 02 = 04

Step 3c) We have now determined two values, 612 and 04 that need to be combined in a very specific way. We now look at our original problem and determine how many more digits our higher factor has over our lower fact. Since our higher factor has 4 digits and our lower factor has 3 digits, there is a 1 digit difference. We now take our intermediate result of 612 and add that number of zeroes to it. We, therefore, add 1 trailing zero to the 612 resulting in 6120. We now add our second result of 04 to this number. This looks like this:


This is the left hand result of our problem. At this point, our solution looks like this:

6124/08 which equals 612,408.

With a little practice, this method will enable you to apply the Base Multiplication method to many more multiplication problems.


Consider 30015 x 716 = (30000 + 15)( 700 + 16 )

= 30000 (700 + 16 ) + 15 x 700 + 15 x 16 .

= 21480000 + 10500 + 240

or more generally ( a + a*)(b +b*) = a (b + b*) + a*b + a*b* .


The German magazine P.M. printed a 5 page article on Vedic Mathematics in their June 2011 edition. The magazine has a print run of 320,000 copies and is distributed in Germany, Austria and Switzerland.
A brief summary (in German) can be seen at: http://www.pm-magazin.de/a/das-kamasutra-der-mathematik


Your comments about this Newsletter are invited.
If you would like to send us details about your work or submit an article or details about a course/talk etc. for inclusion, please let us know on

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Please pass a copy of this Newsletter on to anyone you think may be interested.

Editor: Kenneth Williams

Visit the Vedic Mathematics web site at: http://www.vedicmaths.org

27th June 2011


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