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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?

#### 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!
## Context
>[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
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"](https://math.codidact.com/posts/289079), I loathe that
>[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
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.

#### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k_i ∈ ℕ[]() ∋ \dbinom n { k_1, k_2, \ldots , k_i} = c$ ? For example below, $i = 1, \color{limegreen}{c = 4,072,530}$. Rule out trial and error!
## Context
>[You get two sets of six numbers from 1-45 per $1 play.](https://www.olg.ca/en/lottery/play-lottario-encore/about.html)
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"](https://math.codidact.com/posts/289079), I loathe that
>[2.3 For each 6-number selection chosen, the on-line system shall choose for the player six additional computer-generated
numbers from the class of one to 45.](https://www.olg.ca/content/dam/olg/web/product/rules-game-conditions/lottario-game-conditions.pdf)
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.

combinatorics