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MSC #25 – Counting Events: Labelling Technique

MSC #25 – Counting Events: Labelling Technique

Our 25th Math 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 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. Here we will use the formula nC= n!/ k! (n – k)! =12!/[5!(12 – 5)!]

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

Note that we will get the same result if if 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.

  12C5  7C= 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!) .

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