Verifying Lottery Combinations: 49 Choose 6

[Physics is Phun]

if there is lottery where there are six numbers of 49 drawn and each number is discarded one drawn what are the total combinations.
I think it would be 49 choose 6 but that only gives 14 million and I figured it would be higher. can someone verify this?
thanks

 

[mathman]

I get 13,983,816. It looks like you are right.

 

[chroot]

No, it's not.

It's 49 * 48 * 47 * 46 * 45 * 44 = 10,068,347,520 = 10 billion.

- Warren

 

[jcsd]

The order doesn't matter, so mathman is correct (it's the UK National Lottery I guess).

 

[chroot]

Ah, okay. Silly me.

- Warren

 

[Oggy]

Thats [tex] \left(\begin{array}{cc}49\\6\end{array}\right) [/tex], right?

 

[jcsd]

Yes, that's correct

 

[Physics is Phun]

ok thanks. its a candian lottery Jcsd.
I've got another question not really the same but has to do with choosing so i'll put it here. It was a question I had on a quiz today. Hopefully I got it right
There is a committee to consist of 5 people if there are 15 students and 18 teachers to choose from and the committee needs atleast one student and one teacher how many different committees can be formed.
I believe i did 15c1 * 18c1 * 31c3
is this right?

 

[shmoe]

ok thanks. its a candian lottery Jcsd.
I've got another question not really the same but has to do with choosing so i'll put it here. It was a question I had on a quiz today. Hopefully I got it right
There is a committee to consist of 5 people if there are 15 students and 18 teachers to choose from and the committee needs atleast one student and one teacher how many different committees can be formed.
I believe i did 15c1 * 18c1 * 31c3
is this right?

You're overcounting. If the students are labelled 1, 2, 3,... and the teachers A,B,C,D,.. then the committee 1,A,2,3,4, corresponding to selecting 1 and A as your guaranteed teacher & student and 2, 3, 4 as the 3 chosen from the 31 people left, is counted again as 2,A,1,3,4, where 2 and A are your guaranteed student and teacher and 1,3,4 are from the 31 remaining. (note: the order I've used for the committee is supposed to reflect the reasoning behind your answer)

You could try to correct this overcounting or you could take a different approach. You have 4 acceptable possibilities to make #of students+#of teachers=5, namely 1+4, 2+3, 3+2, and 4+1. Find the number of committees in each case and add.

 

[Alkatran]

Has anyone every calculated the average amount won per lottery ticket? (Including all those little sub prizes). I'd estimate it at max a dollar per ticket, min 10 cents per ticket, probably from 20 to 70.

 

[jcsd]

Has anyone every calculated the average amount won per lottery ticket? (Including all those little sub prizes). I'd estimate it at max a dollar per ticket, min 10 cents per ticket, probably from 20 to 70.

It's actyally a nice little maths project for someone to do at school, I guess that like the UK lottery the expected payout is actually dependent on how many people play the lottery and things like 'roll-overs'.