- **Probability and Statistics**
(*http://mymathforum.com/probability-statistics/*)

- - **How many codes can we create?**
(*http://mymathforum.com/probability-statistics/343683-how-many-codes-can-we-create.html*)

How many codes can we create? |

I'll take a stab at this. 3 places for the dot, $\displaystyle \binom{26}{2}$ ways to choose two letters, $\displaystyle \binom{10}{2}$ ways to choose two numbers, and 4! ways to line them up gives all ways allowing a zero at the start. So now we subtract out all those with zero at the start. Again, 3 places for the dot, $\displaystyle \binom{26}{2}$ ways to choose two letters, 9 ways to choose a digit, and 3! ways to line them up. So, $\displaystyle 3 \cdot \binom{26}{2}\binom{10}{2} \cdot 4! - 3 \cdot \binom{26}{2}\binom{9}{1} \cdot 3! = 100350$ |

All times are GMT -8. The time now is 03:13 PM. |

Copyright © 2018 My Math Forum. All rights reserved.