   SEARCH HOME Math Central Quandaries & Queries  Question from christine, a student: How many six-letter words (not necessarily an English words) are there in which exactly three of the letters are z's? It is a permutation because order matters but it does not matter where the Zs are. Any three spaces can be 1 of 26 letters (because you can repeat letters). The remaining 3 spaces can only be 1 letter. 26^3x 1^3= 17,576 Is this right? No, you answer a different question: "How many six-letter words (not necessarily an English words) are there in which the last three letters are z's?''
There are 26 choices for the first letter, 26 choices for the second, 26 for the third, and only one for the fourth, fifth and sixth, giving 263 x 13. One of the words allowed is zazzzz.

If we switch the question to "How many six-letter words (not necessarily an English words) are there in which the last three letters are z's and there are no other z's?'', then the answer changes to 253 x 13.

If we switch again to your question: "How many six-letter words (not necessarily an English words) are there in which exactly three of the letters are z's?'', then "pizazz'' becomes allowed. It means that the
spots for z's could be any of the {6 choose 3) = 20 choices of three spots among six. How do you work that in the answer?

Claude     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.