Verifying Lottery Combinations: 49 Choose 6

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Discussion Overview

The discussion revolves around calculating the total combinations for a lottery where six numbers are drawn from a set of 49, as well as related combinatorial problems involving committee formation from students and teachers. The scope includes mathematical reasoning and combinatorial calculations.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that the total combinations for a lottery drawing six numbers from 49 is calculated using "49 choose 6," estimating it at 14 million.
  • Another participant calculates the combinations as 13,983,816, agreeing with the initial estimate.
  • A different participant contests the previous calculations, proposing that the total should be 10,068,347,520 based on a different interpretation of the drawing process.
  • There is a clarification that the order of the numbers does not matter, supporting the "choose" calculation.
  • A participant introduces a related question about forming a committee of 5 people from 15 students and 18 teachers, requiring at least one of each group, and presents their calculation method.
  • Another participant points out potential overcounting in the committee calculation and suggests alternative approaches to find the correct number of combinations.
  • Some participants express curiosity about the average winnings per lottery ticket, estimating various amounts based on their assumptions.

Areas of Agreement / Disagreement

Participants express disagreement regarding the total combinations for the lottery, with multiple competing views presented. The committee formation question also remains unresolved, with differing opinions on the correct approach.

Contextual Notes

Participants mention specific lottery types (UK and Canadian) which may influence the interpretation of the drawing process. There is also a discussion about overcounting in combinatorial problems, indicating potential limitations in the initial calculations.

Physics is Phun
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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 :smile:
 
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I get 13,983,816. It looks like you are right.
 
No, it's not.

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

- Warren
 
The order doesn't matter, so mathman is correct (it's the UK National Lottery I guess).
 
Last edited:
Ah, okay. Silly me.

- Warren
 
Thats [tex]\left(\begin{array}{cc}49\\6\end{array}\right)[/tex], right?
 
Yes, that's correct
 
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?
 
Physics is Phun said:
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.
 
  • #10
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.
 
  • #11
Alkatran said:
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'.
 

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