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Need some help with some combinatorics problem! We are supposed to use the Inclusion-Exclusion principle to solve: "how many bridge hands contain exactly 3 clubs, or exactly 5 diamonds or exactly 3 aces?" I know you do something like: |3 clubs|+|3aces|+|5dimonds| - |3clubs AND 3 aces| - |3clubs AND 5 diamonds| -...... +|all 3 intersected| But I'm really struggling with find the cases where the aces may or may not be a club/diamond.

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  • $\begingroup$ Split the three aces into four events: (Ace Club, Ace Diamond, Ace Spade), (Ace Club, Ace Diamond, Ace Heart), (Ace Club, Ace Heart, Ace Spade), (Ace Heart, Ace Diamond, Ace Spade). $\endgroup$ – user348749 Nov 17 '16 at 6:57
  • $\begingroup$ I would start by explaining to the audience how many cards(?) are in a Bridge hand, how many clubs in a deck, how many diamonds in a deck, how many aces in a deck, and how many other types in a deck. We're not professional Bridge players. $\endgroup$ – barak manos Nov 17 '16 at 9:05
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$\begin{align} \text{exactly }3\ \clubsuit &=\underset{\clubsuit}{\binom{13}{3}}\binom{39}{10}=181\,823\,183\,256\\[6pt] \text{exactly }5\ \diamondsuit &=\underset{\diamondsuit}{\binom{13}{5}}\binom{39}{8} =79\,181\,063\,676\\[6pt] \text{exactly }3\ \text{A} &=\underset{\text{A}}{\binom{4}{3}}\binom{48}{10} =26\,162\,863\,584 \end{align}$


$\begin{align} \text{exactly }3\ \clubsuit\text{ and }5\ \diamondsuit &=\underset{\clubsuit}{\binom{13}{3}}\underset{\diamondsuit}{\binom{13}{5}}\binom{26}{5} =24\,212\,433\,960\\[6pt] \text{exactly }3\ \clubsuit\text{ and }3\ \text{A} &=\overbrace{\underset{\clubsuit}{\binom{12}{2}}\underset{\text{A}}{\binom{3}{2}}\binom{36}{8}}^{\text{with A$\clubsuit$}} +\overbrace{\underset{\clubsuit}{\binom{12}{3}}\underset{\text{A}}{\binom{3}{3}}\binom{36}{7}}^{\text{without A$\clubsuit$}} =7\,828\,036\,920\\[6pt] \text{exactly }5\ \diamondsuit\text{ and }3\ \text{A} &=\overbrace{\underset{\diamondsuit}{\binom{12}{4}}\underset{\text{A}}{\binom{3}{2}}\binom{36}{6}}^{\text{with A$\diamondsuit$}} +\overbrace{\underset{\diamondsuit}{\binom{12}{5}}\underset{\text{A}}{\binom{3}{3}}\binom{36}{5}}^{\text{without A$\diamondsuit$}} =3\,191\,048\,784 \end{align}$


$\begin{align} \text{exactly }3\ \clubsuit\text{ and }5\ \diamondsuit\text{ and }3\ \text{A} &=\overbrace{\underset{\clubsuit}{\binom{12}{2}}\underset{\diamondsuit}{\binom{12}{4}}\underset{\text{A}}{\binom{2}{1}}\binom{24}{4}}^{\text{with A}\clubsuit\text{ and A}\diamondsuit} +\overbrace{\underset{\clubsuit}{\binom{12}{2}}\underset{\diamondsuit}{\binom{12}{5}}\underset{\text{A}}{\binom{2}{2}}\binom{24}{3}}^{\text{with A}\clubsuit\text{ without A}\diamondsuit}\\[6pt] &+\overbrace{\underset{\clubsuit}{\binom{12}{3}}\underset{\diamondsuit}{\binom{12}{4}}\underset{\text{A}}{\binom{2}{2}}\binom{24}{3}}^{\text{without A}\clubsuit\text{ with A}\diamondsuit} +\overbrace{\underset{\clubsuit}{\binom{12}{3}}\underset{\diamondsuit}{\binom{12}{5}}\underset{\text{A}}{\color{#C00000}{\binom{2}{3}}}\binom{24}{2}}^{\text{without A}\clubsuit\text{ or A}\diamondsuit}\\[9pt] &=1\,020\,514\,968 \end{align}$


Inclusion-Exclusion says we have $$ \begin{align} &(181\,823\,183\,256+79\,181\,063\,676+26\,162\,863\,584)\\ &-(24\,212\,433\,960+7\,828\,036\,920+3\,191\,048\,784)\\ &+1\,020\,514\,968\\ &=252\,956\,105\,820 \end{align} $$ bridge hands with exactly $3\ \clubsuit$ or exactly $5\ \diamondsuit$ or exactly $3\ \text{A}$.

Since there are $\binom{52}{13}=635\,013\,559\,600$ bridge hands, the probability of the specified hands is approximately $39.834756596275\%$.

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    $\begingroup$ +1, for the formatting and the inclusion of the suit symbols. $\endgroup$ – Willie Wong Nov 17 '16 at 19:12
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HINT: Let’s assemble the pieces. The first four are straightforward, and I’ll just state them without explanation, but feel free to ask about them if necessary.

  • There are $\binom{13}3\binom{39}{10}$ hands with exactly $3$ clubs.
  • There are $\binom{13}5\binom{39}8$ hands with exactly $5$ diamonds.
  • There are $\binom43\binom{48}{10}$ hands with exactly $3$ aces.
  • There are $\binom{13}3\binom{13}5\binom{26}5$ hands with exactly $3$ clubs and exactly $5$ diamonds.

Now we’ll tackle counting the hands with exactly $3$ clubs and exactly $3$ aces. There are two possibilities.

  • If the hand includes the ace of clubs, it must contain exactly $2$ other clubs and exactly $2$ other aces. There are $\binom{12}2\binom32$ ways to pick these $4$ cards and $\binom{36}8$ ways to pick the remaining $8$ cards from the $36$ cards that are neither club nor ace, so there are $\binom{12}2\binom32\binom{36}8$ hands in this case.
  • If the hand does not include the ace of clubs, it must have all $3$ of the other aces. That leaves $10$ cards to be chosen, exactly $3$ of which must be clubs. All $13$ clubs are still available, so there are $\binom{12}3$ ways to choose the $3$ clubs; that leaves $36$ cards that are neither ace nor club, and we must choose $7$ of them, so there are $\binom{12}3\binom{36}7$ hands in this case.

Thus, there are $\binom{12}2\binom32\binom{36}8+\binom{12}3\binom{36}7$ hands with exactly $3$ clubs and exactly $3$ aces.

You can use essentially the same analysis to find the number of hands with exactly $3$ aces and exactly $5$ diamonds.

The final piece needed is the number of hands with exactly $3$ aces, exactly $3$ clubs, and exactly $5$ diamonds. The analysis here works the same way, except that you should break it into four cases:

  • the hand includes both the ace of clubs and the ace of diamonds;
  • the hand includes the ace of clubs but not the ace of diamonds;
  • the hand includes the ace of diamonds but not the ace of clubs;
  • the hand includes neither the ace of clubs nor the ace of diamonds.
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  • $\begingroup$ In the computation of the $3$ clubs and $3$ aces, there are two cases: with and without the ace of clubs. The case with the ace of clubs seems correct. However, I believe the case without the ace of clubs should count $\binom{12}{3}$ ways of choosing $3$ clubs from the clubs minus the ace. Otherwise, I think our counts agree. (+1) $\endgroup$ – robjohn Nov 18 '16 at 2:56
  • $\begingroup$ @robjohn: Yep; hit the wrong key and didn't notice. Thanks! $\endgroup$ – Brian M. Scott Nov 18 '16 at 4:43
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It is easier to calculate directly. Let us define the following events:

$A_S:$ hand contains Ace Club, Ace Diamond, Ace Hearts

$A_C:$ hand contains Ace Spade, Ace Diamond, Ace Hearts

$A_D:$ hand contains Ace Spade, Ace Club, Ace Hearts

$A_H:$ hand contains Ace Spade, Ace Diamond, Ace Club

For $A_S$, the number of hands that also satisfy the other conditions is $$\binom{12}{2}\binom{12}{4}\binom{24}{4}$$ since we need to choose 2 non ace clubs, 4 non ace diamonds and with Ace of heart being present adn Ace of spade excluded, we have 24 cards to choose 4 remaining cards. Similarly for $A_C$, we have $$\binom{12}{3}\binom{12}{4}\binom{24}{3}$$ choices. For $A_D$, the choices are $$\binom{12}{2}\binom{12}{5}\binom{24}{3}$$ and for $A_H$, we have $$\binom{12}{2}\binom{12}{4}\binom{24}{4}$$ choices. We add the above to get the total.

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