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How many combination of 6 numbers are there in 49?

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  • 1 decade ago
    Favourite answer

    There are a total of 13,983,816 different groups of six numbers which could be drawn from the set {1, 2, ... , 49}. To see this we observe that there are 49 possibilities for the first number drawn, following which there are 48 possibilities for the second number, 47 for the third, 46 for the fourth, 45 for the fifth, and 44 for the sixth. If we multiply the numbers 49 x 48 x 47 x 46 x 45 x 44 we get 10,068,347,520. However, each possible group of six numbers (combination) can be drawn in different ways depending on which number in the group was drawn first, which was drawn second, and so on. There are 6 choices for the first, 5 for the second, 4 for the third, 3 for the fourth, 2 for the fifth, and 1 for the sixth. Multiply these numbers out to arrive at 6 x 5 x 4 x 3 x 2 x 1 = 720. We then need to divide 10,068,347,520 by 720 to arrive at the figure 13,983,816 as the number of different groups of six numbers (different picks). Since all numbers are assumed to be equally likely and since the probability of some number being drawn must be one, it follows that each pick of six numbers has a probability of 1/13,983,816 = 0.00000007151. This is roughly the same probability as obtaining 24 heads in succession when flipping a fair coin!

    Source(s): only got it off the internet, not that clever myself... http://www.math.mcmaster.ca/fred/Lotto/
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  • 1 decade ago

    Starting with a bag of 49 differently-numbered lottery balls, there is clearly a 1 in 49 chance of predicting the number of the first ball selected from the bag.

    Accordingly, there are 49 different ways of choosing that first number.

    When the draw comes to the second number, there are now only 48 balls left in the bag (because the balls already drawn are not returned to the bag), so there is now a 1 in 48 chance of predicting this number.

    Thus, each of the 49 ways of choosing the first number has 48 different ways of choosing the second. This means that the odds of correctly predicting 2 numbers drawn from 49 is calculated as 49 × 48.

    On drawing the third number there are only 47 ways of choosing the number; but of course someone picking numbers would have gotten to this point in any of 49 × 48 ways, so the chances of correctly predicting 3 numbers drawn from 49 is calculated as 49 × 48 × 47.

    This continues until the sixth number has been drawn, giving the final calculation, 49 × 48 × 47 × 46 × 45 × 44, which can also be written as :[49! / 49 -6)!]

    This works out to a very large number, 10,068,347,520, which is however much bigger than the 14 million stated above.

    The last step is to understand that the order of the 6 numbers is not significant.

    That is, if a ticket has the numbers 1, 2, 3, 4, 5, and 6, it wins as long as all the numbers 1 through 6 are drawn, no matter what order they come out in.

    Accordingly, given any set of 6 numbers, there are 6 × 5 × 4 × 3 × 2 × 1 = 6! or 720 ways they could be drawn.

    Dividing 10,068,347,520 by 720 gives 13,983,816, (Approx 14 million)

    {also written as [49! / (6! × (49 - 6)!)]}

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  • Nick J
    Lv 4
    1 decade ago

    Easy to work out.

    Choose first number - there are 49 to choose from.

    Choose 2nd number - there are 48 left to choose from

    Choose 3rd number - there are 47 left to choose from

    ....

    choose 6th number - there are 44 to left choose from.

    so the number of different combinations to choose between are 49 x 48 x 47 x 46 x 45 x 44 = 10068347520

    but since it doesn't matter what order you pick them in (in other words the set (1,2,3,4,5,6) is the same as (6,5,4,3,2,1), you must divide that number by the number of ways of ordering 6 numbers. This is 6 x 5 x 4 x 3 x 2 x 1 = 6! = 720

    so, 1006834752 / 720

    This gives the true answer of 13,983,816

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  • 4 years ago

    49 Numbers

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  • 1 decade ago

    calculate 6-49 lotto possible combinations:

    49/6 = 49!/6! = (49x48x47x46x45x44) / (6x5x4x3x2x1)

    = 13,983,816

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  • 5 years ago

    Hello how would I find out combination of numbers that can only ever be lowest to highest and there can be no duplicate numbers?

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  • 5 years ago

    649

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  • 1 decade ago

    (49!)/(6!43!)

    This is read as

    Where x! = (x)* (x-1)*(x-2)*(x-3)*.....*(2)*(1)

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