Question 1144351
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The total number of codes that can be formed of five different symbols that are not repeated in any valid code is the number of permutations of five things taken five at a time, to wit: *[tex \Large 5!\ =\ 5\ \times\ 4\ \times\ 3\ \times\ 2\ \times\ 1] because there are 5 ways to pick the first letter, and for each of those ways, there are 4 ways to pick the second letter making 20 ways to pick the first two letters. Then, for each of those 20 ways, there are 3 ways to pick the third letter, and so on.


The number of codes that start with one particular symbol and ends with a particular symbol is *[tex \Large 1\ \times\ 3\ \times\ 2\ \times\ 1\ \times\ 1].


The probability of any event is the number of successful outcomes divided by the number of total outcomes, or, in this case, the number of codes beginning and ending as specified divided by the total number of possible codes.   
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
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