The math league is holding a lottery fundraiser. Using the number from 1 to 10, participants will choose three different numbers for their lottery ticket. If the order of the numbers does matter (meaning that (1,2,3) and (1,3,2) are different tickets), how many different lottery tickets are possible?
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I don't even know where to start with this problem
- Thread starter twix3474
- Start date
That is a factorial. i.e. 10! means 10*9*8*7*6*5*4*3*2=3628800
Anyway, P(10,3)=720
\(\displaystyle \frac{10!}{7!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2}=10\cdot 9\cdot 8=720\)
Anyway, P(10,3)=720
\(\displaystyle \frac{10!}{7!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2}=10\cdot 9\cdot 8=720\)
D
Deleted member 4993
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twix3474 said:Just for future reference. What change would I make to this formula if I had a similar problem with the only difference being that the order of the numbers did not matter?
Then it becomes a problem of combination. Look up in your textbook for appropriate equation - or do a google search.