# MSC #25 – Counting Events: Labeling Technique

Merry Christmas! 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 between the complicated permutation or combination formulas, he recommends using the word forming formula.

It is quite easy to remember
that since there are two Bs and four Os for a total of six letters in “BOOBOO”, we can form** **
** **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 can be solved using ** ^{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. We will use ^{n}C_{k }= n!/ k! (n – k)! =12!/[5!(12 – 5)!]

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

We
can also label **5** as “first five”
and **7** as “reserves “. We will
arrive at the same solution:
.

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!)** .

More examples would be discussed in Chapter 25 of “30 Master Strategies in Computing”