Mathematics

#10: Post edited by $user avatar$ Chgg Clou‭ · 2023-08-26T04:37:14Z (9 months ago)

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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!
## 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 integers $≥ 32$ to boost my probability of winning](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 $n, k_i ∈ ℕ ∋ \dbinom n { k_1, k_2, ..., k_i} = c$ ? 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.

#9: Post edited by $user avatar$ Chgg Clou‭ · 2023-08-25T23:13:52Z (9 months ago)

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~~Without trial and error, how can I effortlessly deduce all $n, k ∈ ℕ ∋ \binom n {k_1, k_2, ..., k_n} =$ given c?~~

Without trial and error, how can I effortlessly deduce all $n, k_i ∈ ℕ ∋ \binom n {k_1, k_2, ..., k_n} =$ given c?

#8: Post edited by $user avatar$ Chgg Clou‭ · 2023-08-25T23:13:40Z (9 months ago)

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~~Without trial and error, how can I effortlessly deduce all $n, k ∈ ℕ ∋ \binom n {k} =$ given c?~~

Without trial and error, how can I effortlessly deduce all $n, k ∈ ℕ ∋ \binom n {k_1, k_2, ..., k_n} =$ given c?

#### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. 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 integers $≥ 32$ to boost my probability of winning](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 $n, k_i ∈ ℕ ∋ \dbinom n { k_1, k_2, ..., k_i} = c$ ? 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 integers $≥ 32$ to boost my probability of winning](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.

#7: Post edited by $user avatar$ Chgg Clou‭ · 2023-08-25T10:50:12Z (9 months ago)

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#### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. 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 integers $≥ 32$ to boost my probability of winning](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 simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
>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 still buy physical paper tickets.

#### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. 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 integers $≥ 32$ to boost my probability of winning](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.

#6: Post edited by $user avatar$ Chgg Clou‭ · 2023-08-25T10:49:18Z (9 months ago)

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#### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. 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 integers $≥ 32$ to boost my probability of winning](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 simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
>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 still buy physical paper tickets.

#### With online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. 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 integers $≥ 32$ to boost my probability of winning](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 simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
>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 still buy physical paper tickets.

#5: Post edited by $user avatar$ Chgg Clou‭ · 2023-08-25T10:48:59Z (9 months ago)

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#### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. 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 integers $≥ 32$ to boost my probability of winning](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 simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
>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, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already!

#### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. 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 integers $≥ 32$ to boost my probability of winning](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 simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
>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 still buy physical paper tickets.

#4: Post edited by $user avatar$ Chgg Clou‭ · 2023-08-25T10:43:21Z (9 months ago)

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#### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. 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 integers $≥ 32$ to boost my probability of winning](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 simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
>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, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!

#### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. 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 integers $≥ 32$ to boost my probability of winning](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 simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
>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, n to satisfy $\dbinom n {k} =$ any integer around 4 million. If this was possible, our actuaries would have actioned this already!

#3: Post edited by $user avatar$ Chgg Clou‭ · 2023-08-25T10:42:07Z (9 months ago)

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With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ϶ \dbinom n {k} = c$ ? For the example below, $p = 4, 072, 530$. 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$) = \dbinom {45}{6}\color{red}{{/2}} = 4,072,530$.~~
But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](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 OLG why OLG don't simply let players pick this 2nd 6-tuple. The telephone operator replied that~~
>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, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!

#### With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ∋ \dbinom n {k} = c$ ? For example below, $\color{limegreen}{c = 4,072,530}$. 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 integers $≥ 32$ to boost my probability of winning](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 simply let players pick this 2nd 6-tuple. OLG's phone operator replied that
>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, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!

#2: Post edited by $user avatar$ Chgg Clou‭ · 2023-08-25T10:39:45Z (9 months ago)

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~~Without trial and error, how can I efficaciously deduce all $n, k ∈ ℕ ϶ \binom n {k} =$ given c?~~

Without trial and error, how can I effortlessly deduce all $n, k ∈ ℕ ∋ \binom n {k} =$ given c?

~~With an online or computer software, for a given $c ∈ ℕ $, how can I swiftly deduce all natural numbers that $n, k ∈ ℕ ϶ \dbinom n {k} = c$ ? For the example below, $p = 4, 072, 530$.~~
#### 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$) = \dbinom {45}{6}\color{red}{{/2}} = 4,072,530$.
But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](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 OLG why OLG don't simply let players pick this 2nd 6-tuple. The telephone operator replied that
>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, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!

With an online or computer software, for a given $c ∈ ℕ $, how can I efficiently deduce all natural numbers that $n, k ∈ ℕ ϶ \dbinom n {k} = c$ ? For the example below, $p = 4, 072, 530$. 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$) = \dbinom {45}{6}\color{red}{{/2}} = 4,072,530$.
But as [I prefer to pick integers $≥ 32$ to boost my probability of winning](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 OLG why OLG don't simply let players pick this 2nd 6-tuple. The telephone operator replied that
>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, n to satisfy $\dbinom n {k} = 4,072,530$. If this was possible, our actuaries would have actioned this already!

#1: Initial revision by $user avatar$ Chgg Clou‭ · 2023-08-25T10:37:06Z (9 months ago)

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Without trial and error, how can I efficaciously deduce all $n, k ∈ ℕ ϶ \binom n {k} =$ given c?

With an online or computer software, for a given $c ∈ ℕ $, how can I swiftly deduce all natural numbers that $n, k ∈ ℕ ϶ \dbinom n {k} = c$ ? For the example below, $p = 4, 072, 530$. 

#### 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$) = \dbinom {45}{6}\color{red}{{/2}} = 4,072,530$.

But as [I prefer to pick integers $≥  32$ to boost my probability of winning](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 OLG  why OLG don't simply let players pick this 2nd 6-tuple. The telephone operator replied that

>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, n to satisfy $\dbinom n {k} = 4,072,530$.  If this was possible, our actuaries would have actioned this already!

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