## anonymous 2 years ago what is the probability that your five number combination will be the winning combination in a lottery that involves the selection of random numbers from 1 through 20? the order of the five numbers is irrelevant.

1. anonymous

When order matters, the total number of tickets (sample space) is $$20^5$$ We can remove order by dividing out the number of permutations from the sample space. The $$5$$ numbers can be permuted $$5!$$ ways. Thus I ways say the probability is: $\frac {5!}{20^5}$

2. anonymous

@wio can I leave my answer like that or do I have to actually multiply them out?

3. anonymous

You should multiply them out.

4. anonymous

$5!=5\times4\times3\times2\times 1=20\times 3\times 2$

5. anonymous

$\frac{20\times 3\times 2}{ 20^5} = \frac{3\times 2}{20^4} = \frac{3}{10^4\times 2^3}$At this point you may just want to use a calculator.

6. anonymous

this is what i thought it was: 20*19*18*17*16*15! / 15! * 5*4*3*2*1 = 1,860,480 / 120 = 15,504... you can only get one winning ticket so... 1 / 15,504 is that wrong?

7. anonymous

I don't understand the reasoning behind that so I can't tell you what is wrong with it.

8. anonymous

total no:(1-20)=29 use combination ,u have to select only 5 no's total no of ways=20 C 5 =20! / (15! * 5!) =20*19*18*17*16*15! / (15! * 5*4*3*2*1) =15,504 since only one will be correct out of these ways therefore answer is 1/15,504

9. anonymous

I meant to put (1-20)=20! at the top

10. anonymous

Well, the reason we use $$20^5$$ instead of $$20!/15!$$ is because it didn't say we can't have repeated numbers.

11. anonymous

It didn't say we can't chose something like 1 1 1 1 1

12. anonymous

got it... thx