I don't even know where to start with this problem

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twix3474

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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?
 
It's a permutation.

\(\displaystyle P(n,k)=\frac{n!}{(n-k)!}\)

In this case,n=10 and k=3.
 
ok I am still puzzeld. What do the (!) in the formula represent? 10/(10-3)= 10/7 = 1.43. I know that can't be the answer so what steps am I missing or doing incorrectly?
 
That is a factorial. i.e. 10! means 10*9*8*7*6*5*4*3*2=3628800

Anyway, P(10,3)=720

\(\displaystyle \frac{10!}{7!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2}=10\cdot 9\cdot 8=720\)
 
wait I think I remember what the (!) means. That is the symbol for 1x2x3x4 etc. Right?
 
thank you so much for your help. I will try it again. I think I can get it now!
 
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?
 
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?
Click to expand...

Then it becomes a problem of combination. Look up in your textbook for appropriate equation - or do a google search.
 
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