Math Is Fun Forum

Paulm

Guest

Lotto combinations

I have been looking at permitations and combinations section and the information is great but I want to try and take this a step further but I do not know how.  What I am trying to find is how many lottery combinations there would be (given 6 from 49 numbers) where there are no more than 3 numbers in a row of 10 (i.e. between 1-9, 10-19, 20-29, 30-39 and 40-49).  Thanks for your time and hopefully you can help.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Lotto combinations

Consider how many ways there are of having more than 3 numbers in the same row.

(1) All 6 numbers are in the same row.

(2) 5 of the numbers are in the same row and the last number is in another row.

(3) 4 of the numbers are in the same row and the other 2 numbers are in another row (or rows).

Case (1): If the 6 numbers are in the first row, there are

ways of choosing them. If they are in a row other than the first, then there are

ways of choosing them in that row; as there are four possible non-first rows, there are 4 × 210 = 840 ways altogether of choosing them this way. Hence the total number of possibilities for Case (1) is 84 + 840 = 924.

Case (2): If the 5 numbers are in the first row, there are

ways of choosing them, leaving 40 possibilities for the sixth number; so there are 40 × 126 = 5040 ways of choosing the numbers this way. If the 5 numbers are in a row other than the first, then there are

ways of choosing them, leaving 39 possibilities for the sixth number; so there are 39 × 1008 = 39312 ways of choosing the numbers this way. Hence the total number of possibilities for Case (2) is 5040 + 39312 = 44352.

Case (3):
(i) Suppose the 4 numbers in a row are in the first row. There are

ways of choosing them. If the last two numbers are in the same row, there are

ways of choosing those two numbers. If they are in different rows, then there are 6 × 10 × 10 = 600 ways of choosing them. Hence the total number of possibilities for this subcase is 126 × 780 = 98280.

(ii) Suppose the 4 numbers in a row are not in the first row. Then there are

ways of choosing these 4 numbers. Now consider the other two numbers. They can be: (a) both in the first row, (b) both in the same row but not the first, or (c) in different rows. For (a),

. For (b),

. For (c), if one of the numbers is the first row, 9 × (3 × 10) = 270, otherwise 3 × 10 × 10 = 300, giving 270 + 300 = 570. So there are 36 + 135 + 570 = 741 ways of choosing the last 2 numbers, and hence the total number of possibilities for this subcase is 840 × 741 = 622440.

Therefore, the number of possibilities for Case (3) is 98280 + 622440 = 720720.

Thus, the number of possible combinations in which at least four numbers are in the same row is 924 + 44352 + 720720 = 765996.

And the answer to your question is

Last edited by JaneFairfax (2010-03-19 04:02:49)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Lotto combinations

Hi;

Jane does good work!

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Paulm

Guest

Re: Lotto combinations

Many thanks Jane.  I can't believe that there are still so many combinations even though the amount of numbers that are in a row are reduced but I know your calculations are correct.

Say hello to Emma for me.