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# Counting methods

Q.  2-Part Questions!
(a) How many different three-letter initials can people have?
(b) How many different three-letter initials with none of the letters repeated can
people have?

A:

(a)  OK.  We have three-letter initials, that means we have three slots to fill:  __ __ __.

How many choices for the first slot? 26, right? There are 26 letters.  26 __ __.

How many choices for the second slot? 26 again for the 26 different letters:  26 26 __.

And finally, also 26 options for the last slot: 26 26 26.

So, the answer is that there are 26*26*26 = 26³ different three-letter initials.

(b) Now, the same concept, but no repetitions.  Start with 3 blank slots: __ __ __.

How many choices for the 1st slot? 26 different letters means 26 choices: 26 __ __

How many choices for the 2nd slot? Remember, we cannot repeat.  Therefore, there are only 25 choices now:  26 25 __

How many choices for the 3rd and final slot? No repeating means only 24 options now:  26 25 24.

So, there are 26*25*24 different ways.

## One thought on “Counting methods”

1. wow, you make it look very easy for me, i am grasping it more now, thanks 🙂