Al asked in Science & MathematicsMathematics · 1 month ago

how do I calculate the different number of combinations?

theres a lottery with 6 winning numbers and the numbers range from 1-40. Would the possibe combinations be 6^40 or 40^6

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  • 1 month ago

    The possible combination=

    40C6=

    40!/36!6!=

    40*39*38*37*36*35/(6*5*4*3*2*1)=

    10*13*19*37*6*7=

    3838380

  • 1 month ago

    It would be 40^6 if the numbers can be drawn more than once (replacement)

    But it's not the case.  Here's how the logic would work:

    The first number can be any of the 40 numbers.

    The second number can be any of the unused 39 numbers.

    The third number can be any of the unused 38 numbers.

    forth - 37 numbers

    fifth - 36 numbers

    sixth - 35 numbers

    If we multiply these together we get:

    40 * 39 * 38 * 37 * 36 * 35 = 2,763,633,600

    or if you want to use factorial notation:

    40! / 34! = 2,763,633,600

    But this isn't the number as the order the numbers are drawn doesn't matter in lotteries.  In this case, the following sets of numbers count as separate sets:

    1, 2, 3, 4, 5

    2, 3, 4, 5, 1

    5, 4, 3, 2, 1

    Since order doesn't matter we need to take out duplicates like the three sets above.  To do that we divide by 6! (since there are 6 numbers)

    This gives us:

    2763633600 / 6!

    2763633600 / 720

    3,838,380 different combinations of winning numbers in this lottery.

  • 1 month ago

    there are 40 ways to pick #1, 40 ways to pick #2...  40^6=  4,096,000,000.0

  • 1 month ago

    Neither.

    6^40 has no bearing on the question.

    40^6 would be the number of possible sequences if repeats were allowed--you would have 40 choices for each of the six.

    The correct approach works like this:

    You have 40 choices for the first number.

    That can't repeat, so you have only 39 for the next.

    Then 38, 37, 36, 35.

    Multiply those all together and you get 2,763,633,600.

    But you aren't quite done yet, because you have repeats, such as 1 2 3 4 5 6 and 6 5 4 3 2 1. Those are different permutations but the same combination. Each of those are repeated 6! = 720 times, so you need to divide the above number by 720, to get the final answer, 3,838,380.

    That's how you derive the well-known formula for the number of combinations of n objects, taken k at a time:

    nCk = n! / [ (n-k)! k! ]

    n! / (n-k)! give you the number of permutations of n things taken k at a time, and further dividing by k! eliminates the repeats and gives you the number of combinations.

    These are very basic concepts in combinatorics or finite math, so if you are studying this in math, you should learn this very well. It's very important.

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  • 1 month ago

    This is a Combination problem

    ₄₀C₆  = 40! / [ (40-6)! 6! ]

    ₄₀C₆ = 40*39*38*37*36*35 / (6*5*4*3*2*1)

    ₄₀C₆ = 2,763,633,600 / 720

    ₄₀C₆ = 3,838,380 <––––––

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