- #1

ZeroPivot

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## Homework Statement

Lets say a car plate has 6 positions, three of them are numbers and 3 of them are alphabetic.

e.g. ABC 123. I need to figure out how many possible permutations can occur using all the numbers and alphabets BUT i cannot use any number or alfabet twice in an permutation eg. AAB 123 or ABC 112 is forbidden.

## Homework Equations

## The Attempt at a Solution

first i split the problem, first the numbers then the alfabet:

NUMBERS:

so N1 is the number of numbers i can use which is 10 obviously. 0-9

K is the number of positions which is 3.

N1!/(N1-K)=10!/(10-3)!

ALFABET:

so N2 is the number of alfabets which is 26, K is the number of positions 3.

N2!/(N2-K)!=26!/(26-3)!

The ANSWER:

(N1!/(N1-K))*(N2!/(N2-K)!) = (10!/(10-3)!)*(26!/(26-3)!)

is the answer correct?