Finding Number of 3-Letter Codes

Algebra 2 · Axiom Academy

EXAMPLE Finding the Number of 3-Letter Codes Applying the Fundamental Counting Principle to count arrangements of letters A security code is made up of 3 letters, each chosen from A to Z. Letters may repeat (for example, "AAA" is a valid code). How many different 3-letter codes are possible? Nice work — you counted arrangements using the Fundamental Counting Principle. Remember: The Fundamental Counting Principle: if one choice can be made in m ways and a second (independent) choice in n ways, the two together can be made in ways — extend to as many positions as needed. Repetition allowed means every position is independent: each of the 3 letters can be any of the 26 letters, even reusing a letter already chosen. Multiply, don't add: the counts for each position multiply together, , not 26+26+26 . Repetition changes the count: if no letter could repeat, the choices would shrink at each position ( ) — that's the permutations formula from earlier in this unit, P(26,3) . The answer: with repetition allowed, there are 26^3=17 , 576 possible 3-letter codes. This same idea — multiply the number of choices at each independent step — works for counting passwords, license plates, phone numbers, and any arrangement built one position at a time.

This is the written version of the interactive lesson above. See the full Algebra 2 course.