**MSC #25: Counting Events – Labelling Technique**

“How many rectangles of all types are there? A) 34; B) 56; C) 289; D) 324; E) 360

This question, given in Intermediate, Senior and Open Levels of the 2^{nd} International Vedic Mathematics Olympiad, can be easily solved using the 25^{th} of our 30 MATH-Inic Specials for Christmas series.

Our 25^{th} MATH-Inic Special for Christmas is not of Vedic Math origin, but it definitely adheres to the Vedic ideal of “at-sight-one-line-mental-solution”.

I learned this technique from the video lessons of James Tanton, the resident mathematician at the Mathematical Association of America.

Instead of deciding which of the complicated permutation or combination formulas to use, he recommends using the word-forming formula.

This word -forming formula is quite easy to remember and apply. In a **six** letter word like “BOOBOO” where we have **two Bs** and **four Os**, we can form** 6!/ (2! 4!) = 15** different words.

Now if we have a basketball team composed of 12 very versatile players who can play in any position, in how many ways can a starting five of a center, a point guard, a shooting guard, a small forward and a power forward be selected?

Using the **multiplication principle**, there are 12 who can be selected as center, and once the center is chosen, there are 11 remaining aspirants for the point guard position and so on, so that there will be **12 x 11 x 10 x 9 x 8 = 95,040** ways.

This is also a **permutation problem **which we usually solve using the formula ^{n}**P _{k} = n!/ (n – k)! = 12! / (12 – 5)! **=

**12!/7!**

**= (12 x 11 x 10 x 9 x 8 x 7!)/7! = 95,040**

But we can just label **1** as center, **1** as point guard, **1** as shooting guard, **1** as short forward, **1** as power forward and **7** as unselected players and use the word forming formula, **12!**/**(1!1!1!1!1!7!) = 12!/7! **= **95,040.**

If we just want to choose any 5 players to form the starting five, this will be a **combination problem **of 12 players taken 5 at a time. Here we will use the formula ^{n}**C _{k }= n!/ k! (n – k)! =12!/[5!(12 – 5)!]**

**= 12!/(5! 7!) = 792 ways**

Note that we will get the same result if we want to know the number of possible ways we can chose the 7 reserves, that is, **k = 7**

In any case, we can just label **5 as “first five”**, **7 as “reserves”,** use the word forming formula and arrive at the same solution.

Now if we are to form two five-man teams from a group of 12 players, using the traditional method, we would have to choose 5 players out of 12 AND then choose 5 players out of 7.

^{12}**C _{5 }x ^{7}C_{5 }= 12!/[5! (12 – 5)!] x 7!/[5!(7– 5)!] **=

**12!/(5!7!)**

**x 7!/(5!2!) = 12!/(5!5!2!).**

Or we could just label **5 as “A team**”, **5 as “B team**” and **2 “reserves”** and immediately compute as **12!/(5!5!2!)**.

As for the IVMO question of finding the number of rectangles in the figure given at the start of this article, let us first examine an easier question given in the Junior Level of the same competition:

” How many rectangles of all types are in this grid?

A) 21; B) 22; C) 126; D) 167; E) 168 “

To form a rectangle, we need two horizontal AND two vertical sides. Therefore, to solve this type of problem, we need to know the number of possible ways to choose **2** from **h** horizontal lines and multiply it by the number of ways of choosing **2** from **v** vertical lines.

Using our word forming formula, we can designate the 2 chosen lines as **C**s and those not chosen as **Ns. **Since **C **is always 2 in this case, we will have:

Ways horizontal = h!/[2! (h – 2)!] and ways vertical = v!/[2! (v – 2)!]

For this problem, we have **h = 4** and **v = 8**, therefore, the number of rectangles is 4!/[2! (4 – 2)!] x 8!/[2! (8 – 2)!] = 4!/(2!2!) x 8!/(2!6!) = 6 x 28 = **168**.

Now the question in the higher levels of the IVMO is more complicated:

The composite figure is L shaped, not rectangular so we cannot directly use our formula. However, we can form two rectangles from the figure:

But the rectangles in this section is counted twice:

So we must add the number of rectangles in the first two figures and deduct the number in the 3^{rd} to get the final answer.

1^{st} figure: 9!/[2! (9 – 2)!] x 3!/[2! (3 – 2)!] = 9!/(2!7!) x 3!/(2!1!) = 36 x 3 = **108**.

2nd figure: 7!/[2! (7 – 2)!] x 6!/[2! (6 – 2)!] = 7!/(2!5!) x 6!/(2!4!) = 21 x 15 = **305**.

3^{rd} figure: 7!/[2! (7 – 2)!] x 3!/[2! (3 – 2)!] = 7!/(2!5!) x 3!/(2!1!) = 21 x 3 = **63**.

Total is 108 + 315 – 63 = **360** rectangles.