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There are four qualities that hands may have. There are hands that miss clubs, hands that have no hearts, hands without spades and hands without diamonds.

The situation is tricky because the binomial coefficients intervene in a double way. To reveal this let us use $N(x)$ and $E(x)$ the generating functions for the "hands having at least x qualities" and "hands having exactly x qualities".

Shortly, P.I.E says that $N(x)= E(x+1)$

Here

$N(x) = \binom 4 3 \binom 8 6 x^3 + \binom 4 2 \binom {16} 6 x^2 + \binom 4 1 \binom {24} 6x + \binom 4 0 \binom {32} 6$

$N(x) =112x^3 + 48048x^2 + 538384x + 906192$

$E(x) = N(x-1) = 12x^3 + 47712x^2 + 442624x + 415744$

As one may see, $442624$, the desired number, is

$442664 = \binom 3 1 \binom 4 3 \binom 8 6 - \binom 2 1 \binom 4 2 \binom {16} 6 + \binom 1 1 \binom 4 1 \binom {24} 6$

Suppose now we have a 3-color deck of 24 cards and we are interested in all-color hands. Again,

$N(x) = \binom 3 2 \binom 8 6 x^2 + \binom 3 1 \binom {16} 6 x + \binom 3 0 \binom {24} 6 $

$N(x) =84x^2 + 24024x + 134596$

$E(x) = N(x-1) = 84x^2 + 23846x + 110656$

Again, as we may see, $442624 = 4\times (134596 - 24024 + 84) $

=====

:) And do not be afraid of "generating functions"; they are neither generating nor functions but just mnemonical expressions that fits and help to shorten the writing.

There are four qualities that hands may have. There are hands that miss clubs, hands that have no hearts, hands without spades and hands without diamonds.

The situation is tricky because the binomial coefficients intervene in a double way. To reveal this let us use $N(x)$ and $E(x)$ the generating functions for the "hands having at least x qualities" and "hands having exactly x qualities".

Shortly, P.I.E says that $N(x)= E(x+1)$

Here

$N(x) = \binom 4 3 \binom 8 6 x^3 + \binom 4 2 \binom {16} 6 x^2 + \binom 4 1 \binom {24} 6x + \binom 4 0 \binom {32} 6$

$N(x) =112x^3 + 48048x^2 + 538384x + 906192$

$E(x) = N(x-1) = 12x^3 + 47712x^2 + 442624x + 415744$

As one may see, $442624$, the desired number, is

$442664 = \color {red}{\binom 3 1 }\binom 4 3 \binom 8 6 - \color{red}{\binom 2 1 } \binom 4 2 \binom {16} 6 + \color{red}{\binom 1 1 } \binom 4 1 \binom {24} 6$

Suppose now we have a 3-color deck of 24 cards and we are interested in all-color hands. Again,

$N(x) = \binom 3 2 \binom 8 6 x^2 + \binom 3 1 \binom {16} 6 x + \binom 3 0 \binom {24} 6 $

$N(x) =84x^2 + 24024x + 134596$

$E(x) = N(x-1) = 84x^2 + 23846x + 110656$

$110656 = \color {red}{\binom 2 0 }\binom 3 2 \binom 8 6 - \color{red}{\binom 1 0 } \binom 3 1 \binom {16} 6 + \color{red}{\binom 0 0 } \binom 3 0 \binom {24} 6$

Again, as we may see, $442624 = 4\times 110656$

=====

:) And do not be afraid of "generating functions"; they are neither generating nor functions but just mnemonical expressions that fit and help to shorten the writing.

There are four qualities that hands may have. There are hands that miss clubs, hands that have no hearts, hands without spades and hands without diamonds.

The situation is tricky because the binomial coefficients intervene in a double way. To reveal this let us use $N(x)$ and $E(x)$ the generating functions for the "hands having at least x qualities" and "hands having exactly x qualities".

Shortly, P.I.E says that $N(x)= E(x+1)$

Here

$N(x) = \binom 4 3 \binom 8 6 x^3 + \binom 4 2 \binom {16} 6 x^2 + \binom 4 1 \binom {24} 6x + \binom 4 0 \binom {32} 6$

$N(x) =112x^3 + 48048x^2 + 538384x + 906192$

$E(x) = N(x-1) = 12x^3 + 47712x^2 + 442624x + 415744$

As one may see, $442624$, the desired number, is

$442664 = \binom 3 1 \binom 4 3 \binom 8 6 - \binom 2 1 \binom 4 2 \binom {16} 6 + \binom 1 1 \binom 4 1 \binom {24} 6$

Suppose now we have a 3-color deck of 24 cards and we are interested in all-color hands. Again,

$N(x) = \binom 3 2 \binom 8 6 x^2 + \binom 3 1 \binom {16} 6 x + \binom 3 0 \binom {24} 6 $

$N(x) =84x^2 + 24024x + 134596$

$E(x) = N(x-1) = 84x^2 + 23846x + 110656$

Again, as we may see, $442624 = 4\times (134596 - 24024 + 84) $

There are four qualities that hands may have. There are hands that miss clubs, hands that have no hearts, hands without spades and hands without diamonds.

The situation is tricky because the binomial coefficients intervene in a double way. To reveal this let us use $N(x)$ and $E(x)$ the generating functions for the "hands having at least x qualities" and "hands having exactly x qualities".

Shortly, P.I.E says that $N(x)= E(x+1)$

Here

$N(x) = \binom 4 3 \binom 8 6 x^3 + \binom 4 2 \binom {16} 6 x^2 + \binom 4 1 \binom {24} 6x + \binom 4 0 \binom {32} 6$

$N(x) =112x^3 + 48048x^2 + 538384x + 906192$

$E(x) = N(x-1) = 12x^3 + 47712x^2 + 442624x + 415744$

As one may see, $442624$, the desired number, is

$442664 = \binom 3 1 \binom 4 3 \binom 8 6 - \binom 2 1 \binom 4 2 \binom {16} 6 + \binom 1 1 \binom 4 1 \binom {24} 6$

Suppose now we have a 3-color deck of 24 cards and we are interested in all-color hands. Again,

$N(x) = \binom 3 2 \binom 8 6 x^2 + \binom 3 1 \binom {16} 6 x + \binom 3 0 \binom {24} 6 $

$N(x) =84x^2 + 24024x + 134596$

$E(x) = N(x-1) = 84x^2 + 23846x + 110656$

Again, as we may see, $442624 = 4\times (134596 - 24024 + 84) $

=====

:) And do not be afraid of "generating functions"; they are neither generating nor functions but just mnemonical expressions that fits and help to shorten the writing.

There are four qualities that hands may have. There are hands that miss clubs, hands that have no hearts, hands without spades and hands without diamonds.

The situation is tricky because the binomial coefficients intervene in a double way. To reveal this let us use $N(x)$ and $E(x)$ the generating functions for the "hands having at least x qualities" and "hands having exactly x qualities".

Shortly, P.I.E says that $N(x)= E(x+1)$

Here

$N(x) = \binom 4 3 \binom 8 6 x^3 + \binom 4 2 \binom {16} 6 x^2 + \binom 4 1 \binom {24} 6x + \binom 4 0 \binom {32} 6$

$N(x) =112x^3 + 48048x^2 + 538384x + 906192$

$E(x) = N(x-1) = 12x^3 + 47712x^2 + 442624x + 415744$

As one may see, $442624$, the desired number, is

$442664 = \binom 3 1 \binom 4 3 \binom 8 6 - \binom 2 1 \binom 4 2 \binom {16} 6 + \binom 1 1 \binom 4 1 \binom {24} 6$

There are four qualities that hands may have. There are hands that miss clubs, hands that have no hearts, hands without spades and hands without diamonds.

The situation is tricky because the binomial coefficients intervene in a double way. To reveal this let us use $N(x)$ and $E(x)$ the generating functions for the "hands having at least x qualities" and "hands having exactly x qualities".

Shortly, P.I.E says that $N(x)= E(x+1)$

Here

$N(x) = \binom 4 3 \binom 8 6 x^3 + \binom 4 2 \binom {16} 6 x^2 + \binom 4 1 \binom {24} 6x + \binom 4 0 \binom {32} 6$

$N(x) =112x^3 + 48048x^2 + 538384x + 906192$

$E(x) = N(x-1) = 12x^3 + 47712x^2 + 442624x + 415744$

As one may see, $442624$, the desired number, is

$442664 = \binom 3 1 \binom 4 3 \binom 8 6 - \binom 2 1 \binom 4 2 \binom {16} 6 + \binom 1 1 \binom 4 1 \binom {24} 6$

Suppose now we have a 3-color deck of 24 cards and we are interested in all-color hands. Again,

$N(x) = \binom 3 2 \binom 8 6 x^2 + \binom 3 1 \binom {16} 6 x + \binom 3 0 \binom {24} 6 $

$N(x) =84x^2 + 24024x + 134596$

$E(x) = N(x-1) = 84x^2 + 23846x + 110656$

Again, as we may see, $442624 = 4\times (134596 - 24024 + 84) $