Comments on Without trial and error, how can I effortlessly deduce all $n, k_i ∈ ℕ ∋ \binom n {k_1, k_2, ..., k_n} =$ given c?
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Without trial and error, how can I effortlessly deduce all $n, k_i ∈ ℕ ∋ \binom n {k_1, k_2, ..., k_n} =$ given c?
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With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k_i ∈ ℕ ∋ \dbinom n { k_1, k_2, ..., k_i} = c$ ? For example below, $i = 1, \color{limegreen}{c = 4,072,530}$. Rule out trial and error!
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Hence, $\Pr($winning Lottario's jackpot$) = \dbinom {45}{6}{\color{red}{/2}} = \color{limegreen}{4,072,530}$.
But as I prefer to pick unpopular "numbers to reduce the number of ways you split the prize", I loathe that
I asked why OLG don't let players pick this 2nd 6-tuple. OLG's phone operator replied
I am not an actuary, but I understand what you want. If I recall my high school math, it is impossible for any integers k ≤ 20, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already! OLG requires k ≤ 10, because research proves that players dislike picking over 10 integers, which they find inconvenient. Remember, many players buy physical paper tickets.
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