how shall i work out lottery combination from numbers 1 to 49?
435773
http://mathforum.org/library/drmath/view/56122.html\nhttp://www.ph.tn.tudelft.nl/~ed/pdfs/Gambo-patent.pdf\nhttp://www.math.mcmaster.ca/fred/Lotto/\nGOOD LUCK :)
315772
You simply type: \n=rand(25,25) and then enter.
1065918
To get your answers, imagine what has to happen.  \n\na) First, you have to pick 1 of the 4 white balls out of the 12 balls there.\n\n2nd, you have to pick 1 of the 3 white balls left out of the 11 balls there.\n\nSo, in order to get a red and then a red, you need to get 4 out of 12 followed by 3 out of 11 or 4/12 times 3/11 or 12/132 = 1/22.\n\nb) First, how many ways can 5 people line up?  5!.  How many ways are correct (assuming they all have different names, only 1).  So, the prob is 1/5! or 1/120.
499010
You mean you are taking 20 numbers from 70?\n\nIf that's the case then the number of ways of drawing 20 from 70 is enormous = 161884603662657900\nThis number is too large for a hand calculator\nI've given you the formula below but you can go to this website and just plug numbers in that interest you, whether it's for lottery combinations, or whatever. You'll see the combinations calculator(3rd one down) just plug 70 in first and then 20 in the 2nd space provided and press the equals sign and you'll get your answer.\n\nhttp://fclass.vaniercollege.qc.ca/web/mathematics/real/Calculators/PermsCombs_Calc_1.htm\n\nThe formula used is nCr = n!/(n-r)!r! where n=70 and r=20\nn! = n(n-1)(n-2)x...x4x3x2x1
353352
The answer is Choose(n+k-1,k-1).\n\nFor example, if n=5 and k=3, the value is Choose(7,2) = 21.\n\nOne way to understand that is to take n+k-1 different pigeon holes:\n\n    O O O O O O O\n\nNow, if we choose 2 of these, say, the fourth and the sixth, like:\n    O O O X O X O\n\nthat yields the distribution as (3,1,1).  You can, I hope, see how this works in general.
812879
lucky numbers 79 53 72 92 91 92
11678
Easy! Just predict next week's lottery numbers...\nPS let me know too...
514377
There are 20! possible orderings for picking the numbers from the hat. Only one has them in order, so 1/20!.\n\nAnother way to look at it is that when there are 20 numbers in the hat, your chances are 1/20 of picking the 1. Then they are 1/19 of picking the 2, then 1/18 of picking the 3 etc.\n\nSo the chances are (1/20)(1/19)(1/18)...(1/2)(1/1)\n\nIf you threw the numbers back in each time, the probablity of pulling them out in order would be 1/(20^20).
322033
444? Next weeks lottery numbers! I better write them down!!!
779807
Yiu would first decide how many decimal places you are going to consider.  The usual is 3.  Then work it from there...;
340631
First you must consider that the first digit can not be zero, otherwise it would not be a 6 digit number.\nThen you must consider how you can arrange odd or even within the digits. (O=ODD, E-EVEN)\n1.OOOEEE\n2.OOEOEE\n3.OOEEOE\n4.OOEEEO\n5.OEOOEE\n6.OEOEOE\n7.OEOEEO\n8.OEEOOE\n9.OEEOEO\n10.OEEEOO\n11.EOOOEE\n12.EOOEOE\n13.EOOEEO\n14.EOEOOE\n15.EOEOEO\n16.EOEEOO\n17.EEOOOE\n18.EEOOEO\n19.EEOEOO\n20.EEEOOO\nFor each of the #1-10 above we can consider 5 possible digits for the odds and 5 for the evens\nso for each there are 5*5*5*5*5*5=15625 different digit combinations.\nFor #11-20 we must consider that the first digit can not be zero, so there are 4*5*5*5*5*5=12500 different digit combinations for each.\nSo there is a total of 10*15625 + 10*12500=281250 different numbers with 3 odd and 3 even digits.
908589
Well, unfortunately, in order to have a good chance of winning the big prize, you'll probably have to spend more than the prize itself.  \n\nTo calculate the odds of winning, we start by finding the odds of getting one number.  In your scenario, it seems that there are 80 numbers, of which you've picked 40.  Neither the actual numbers themselves, nor the order they are selected makes any difference, since, presumably, each number has an equal chance of being picked.  For the first number, our odds are 40/80, or 1/2, since we have picked 40 numbers out of 80 total.  Not too bad, huh?\n\nWell, here's where it gets rough in a hurry.  We now need to get the second number.  The odds of us having this number have changed slightly, since one number is now off the board, to 39/79 (we have 39 numbers left, the board has 79 left).  This number must then be multiplied to the previous number, as only half of the situations have us selecting the first number correctly.  This comes out to 0.2468, or a little less than 1/4.\n\nContinuing along, each successive number has slightly decreasing odds, and must also be multiplied by the earlier odds as well.  So the odds on the third number are 40/80 * 39/79 * 38/78.  And so on, until we've calculated the odds of picking the 20th number (21/61) and multiplied it to the other 19 values.\n\nThe final value is a miniscule .00000003899, or 1 in 25.7 million.  By purchasing the second ticket, your odds increase to 2 in 25.7 million, with three tickets, the odds are 3 in 25.7 million, etc.  So to even have half a chance of winning, you will need to have a staggering 12.8 million tickets.  This may seem ridiculously high, but it's how lotteries, kenos, and other similar games of chance are able to do so well.\n\nHope this helps.
126757
45\nYou need a number that is the 2 power of a whole number and falls in that range.
306637
There's a trick to figuring this out quickly.\n\nFor each number between 1 and 499 there is a corresponding number between 501 and 999 which, when added together, equals 1000.\n\nFor example:\n\n1 + 999 = 1000\n2 + 998 = 1000\n...\n499 + 501 = 1000\n\nNow, there are 499 pairs like that.  Add them together and you get 499,000\n\nAdd to that 1000 and 500, and you get 500,500, which is your total.
50282
win the lottery several times
258297
Each game you have a 50/50 chance of being right, so that's 1/2^1(^is the sign for an exponent), there are 32 first round games, so for picking the perfect first round the chance is 1/2^32, for the second there are 16 games, you multiply the odds of picking these right (1/2^16) and continute for the number of games each round, so your formula will be, 1/(2^32*2^16*2^8*2^4*2^2*2^1) which will equal 1/9,223,372,036,854,775,808.  There is 9,223,372,036,854,775,808 different combinations of winners, and only one is correct for the perfect bracket.
838732
000000 to 999999 --> 1 million combinations\n\n10 choices for the first digit\n10 choices for the second digit\n10 choices for the third digit\n10 choices for the fourth digit\n10 choices for the fifth digit\n10 choices for the sixth digit\n\n10 x 10 x 10 x 10 x 10 x 10 = 1 million\n\nHowever, if you disallow 0 as a starting digit then you have the following:\n\n*9* choices for the first digit (1-9)\n10 choices for the second digit (0-9)\n10 choices for the third digit (0-9)\n10 choices for the fourth digit (0-9)\n10 choices for the fifth digit (0-9)\n10 choices for the sixth digit (0-9)\n\n9 x 10 x 10 x 10 x 10 x 10 = 900,000