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# Ok so say I have 40 numbers set as 4 lines of 10...

Topic closed. 5 replies. Last post 1 year ago by SergeM.

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New Member
brisbane
Australia
Member #168511
September 3, 2015
8 Posts
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 Posted: September 20, 2015, 10:17 pm - IP Logged

Ok so say I have 40 numbers set as 4 lines of 10...

(1-10,11-20,21-30 and 31-40). How do we work out every combination possible to get 8 numbers while using a maximum of 5 numbers from each of the 4 lines. As 8 numbers will be chosen this means that on average 2 numbers from each set of 10 will be used to make up the 8 so if we set a maximum of 5 from each set of 10 then we should have it covered. I need to know every possible combination so what I mean is usually if I were to do it manually with all 40 numbers it will be 1,2,3,4,5,6,7,8 then 1,2,3,4,5,6,7,9 then 1,2,3,4,5,6,7,10... all the way to 1,2,3,4,5,6,7,40 then start again with 1,2,3,4,5,6,8,9 all the way to 1,2,3,4,5,6,8,40 and so on but what I want to know is how can we do it and how many possiblitlies will it take if we had a maximum of 5 to choose from in each line such as 1,2,3,4,5,11,12,13.... thank you in advance for your answers
Dallas, Texas
United States
Member #4549
May 2, 2004
1691 Posts
Offline
 Posted: September 22, 2015, 5:43 am - IP Logged

Ok so say I have 40 numbers set as 4 lines of 10...

(1-10,11-20,21-30 and 31-40). How do we work out every combination possible to get 8 numbers while using a maximum of 5 numbers from each of the 4 lines. As 8 numbers will be chosen this means that on average 2 numbers from each set of 10 will be used to make up the 8 so if we set a maximum of 5 from each set of 10 then we should have it covered. I need to know every possible combination so what I mean is usually if I were to do it manually with all 40 numbers it will be 1,2,3,4,5,6,7,8 then 1,2,3,4,5,6,7,9 then 1,2,3,4,5,6,7,10... all the way to 1,2,3,4,5,6,7,40 then start again with 1,2,3,4,5,6,8,9 all the way to 1,2,3,4,5,6,8,40 and so on but what I want to know is how can we do it and how many possiblitlies will it take if we had a maximum of 5 to choose from in each line such as 1,2,3,4,5,11,12,13.... thank you in advance for your answers

The biggest reason people are hesitant to respond to post like this is because they aren't sure what you want.

Can't blame them.

You have 4 lines of ten numbers. Got it.

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

Now what do you want?

Do you want every number combination out of ten?

Do you want every 8 number combinations out of 20 (which would be 5 numbers from each of the four lines)?

Or do you want to know how many 8 number combinations is in 40 numbers?

Or something else?

All the confusing garbage 4 lines/ten numbers....all 8 numbers combinations.....5 numbers max......leaves people scratching their head. Its meaningless dribble.

Its easy to find the answer once we know exactly what you need. Its not easy and most often a pain when we take the time to answer only to find out what you wrote is not what you meant.

Although some will guess and some will think they know, nobody can read your mind.

My greatest accomplishment is teaching cats about Vienna Sausage. When I need a friend, all I need do is walk outside, pop open a can, and every little critter in the neighborhood drops by to say "Hi!"

New Member
brisbane
Australia
Member #168511
September 3, 2015
8 Posts
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 Posted: September 22, 2015, 8:42 pm - IP Logged

Hi mate this is what I mean:

I have the number 1-40 set as a normal keno board like so

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40

I want to know how many sets of 8 numbers I can get from these 40 numbers. The conditions I wish to add to this to cut down the possibilities or bad numbers are:

at least 1 numbers from each line must be use and maximum of 5 numbers from each line will be used.

so:

1,2,3,4,5,6,7,8 is wrong because it uses all numbers from 1 line so it doesn't fulfil the min and max of my conditions

1,2,3,4,5,11,21,31 is correct because it fulfills the min and max of my conditions (5 max from line 1 and 1 min from line 2,3 and 4).

Do you understand now?

Sorry if I didn't explain it better earlier.

If you can help to tell me how many possible options and how to find or write the options in a sequence that would be great. Thanks.

Dallas, Texas
United States
Member #4549
May 2, 2004
1691 Posts
Offline
 Posted: September 22, 2015, 9:37 pm - IP Logged

Hi mate this is what I mean:

I have the number 1-40 set as a normal keno board like so

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40

I want to know how many sets of 8 numbers I can get from these 40 numbers. The conditions I wish to add to this to cut down the possibilities or bad numbers are:

at least 1 numbers from each line must be use and maximum of 5 numbers from each line will be used.

so:

1,2,3,4,5,6,7,8 is wrong because it uses all numbers from 1 line so it doesn't fulfil the min and max of my conditions

1,2,3,4,5,11,21,31 is correct because it fulfills the min and max of my conditions (5 max from line 1 and 1 min from line 2,3 and 4).

Do you understand now?

Sorry if I didn't explain it better earlier.

If you can help to tell me how many possible options and how to find or write the options in a sequence that would be great. Thanks.

Taking any 5 numbers from the first line and combining it with any 5 number from the second line in sets of 8, lowest to highest, no repeating number in any line we get 45 sets.This is true for all 8 number combinations made with any 10 numbers.

The more numbers you add to the total, the trickier it gets.

Taking 5 from line 1, 5 from line 2, 5 from line 3 and 5 from line 4 we have 20 numbers total.

Using any 20 numbers, lowest to highest, no repeats, in combinations of 8 we get 125,970 sets.

The formula is (20*19*18*17*16*15*14*13)/(8*7*6*5*4*3*2*1) = 125,970 sets.

For 40 numbers, lowest to highest, no repeats, in combinations of 8 you get 76,904,685 sets.

Or (40*39*38*37*36*35*34*33)/(8*7*6*5*4*3*2*1) = 76904685.

Hope this helps.

My greatest accomplishment is teaching cats about Vienna Sausage. When I need a friend, all I need do is walk outside, pop open a can, and every little critter in the neighborhood drops by to say "Hi!"

mid-Ohio
United States
Member #9
March 24, 2001
19826 Posts
Offline
 Posted: September 30, 2015, 11:35 am - IP Logged

Hi mate this is what I mean:

I have the number 1-40 set as a normal keno board like so

1 2 3 4 5 6 7 8 9 10

11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30

31 32 33 34 35 36 37 38 39 40

I want to know how many sets of 8 numbers I can get from these 40 numbers. The conditions I wish to add to this to cut down the possibilities or bad numbers are:

at least 1 numbers from each line must be use and maximum of 5 numbers from each line will be used.

so:

1,2,3,4,5,6,7,8 is wrong because it uses all numbers from 1 line so it doesn't fulfil the min and max of my conditions

1,2,3,4,5,11,21,31 is correct because it fulfills the min and max of my conditions (5 max from line 1 and 1 min from line 2,3 and 4).

Do you understand now?

Sorry if I didn't explain it better earlier.

If you can help to tell me how many possible options and how to find or write the options in a sequence that would be great. Thanks.

"I want to know how many sets of 8 numbers I can get from these 40 numbers."

There are 76,904,685 possible combinations of 8 in a pool of 40 numbers.

* you don't need to buy more tickets, just buy a winning ticket *

Economy class
Belgium
Member #123700
February 27, 2012
4035 Posts
Offline
 Posted: September 30, 2015, 12:23 pm - IP Logged

With regular 20/70, playing 40 numbers is over the top. You can consider it as a wheel. Best score you can reach is 20/40 and the worst score is 0/40. Playing 20 numbers can end in 20/20 to 0/20. That is all there is to say about this.

Getting 20/40, you might just get four times 5/10 by Murphy's law, where all goes wrong anyway.

I did a quickpick version with 70 numbers, randomnly spread over 7 combinations, on my system page.

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