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MSC #25 – Counting Events: Labeling Technique

MSC #25 – Counting Events: Labeling Technique

Merry Christmas! Our 25th 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 nPk = 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 nCk = 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.

  12C5  x 7C5 = 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”

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