{36 \choose 6} = \frac{36!}{6!30!} = \frac{36 \cdot 35 \cdot 34 \cdot 33 \cdot 32 \cdot 31}{6!} = \frac{P(36, 6)}{6!}. Any of these answers, or anything equivalent to them, are correct.

There are 36 cards, and we must choose of 6 of them at random without replacement such that order matters.

{12 \choose 6} \cdot 3^6. Anything equivalent to this answer is also correct.

Since we’re asking for the number of hands with unique types, we first pick 6 different types, which can be done {12 \choose 6} ways. Then, for each chosen type, there are {3 \choose 1} = 3 ways of selecting the color. So, the colors can be chosen in 3 \cdot 3 \cdot ... \cdot 3 = 3^6 ways.

{12 \choose 2} = 66. Anything equivalent to this answer is also correct.

Once we pick 2 unique plane types, we need to choose 3 different colors per type, out of a set of 3 possibilities. This can be done, for each type, in {3 \choose 3} = 1 way.

{12 \choose 3} \cdot {3 \choose 2}^3 = {12 \choose 3} 3^3. Anything equivalent to this answer is also correct.

First, we choose 3 distinct types, which can be done in ${12 \choose 3} ways. Then, for each type, we choose 2 colors, which can be done in {3 \choose 2} ways per type.

{12 \choose 3} \cdot {3 \choose 2}^3 + {12 \choose 1}{11 \choose 1}{3 \choose 2}{10 \choose 1}{3 \choose 1}. Anything equivalent to this answer is also correct.

If we select 6 cards and end up with 3 unique plane types, there are two possibilities: - We get 2 of each type (e.g. xx yy zz). This is the case in part (d), so the total number of sets of 6 cards that fall here are {12 \choose 3} \cdot {3 \choose 2}^3. - We get 3 of one type, 2 of another, and 1 of the last one (e.g. xxx yy z). - xxx: There are 12 options for the type that appears 3 times, so that can be selected in {12 \choose 1} ways. We must take all 3 of that types colors, which can be done in {3 \choose 3} = 1 way. - yy: There are now 11 options for the type that appears 2 times, and for that type, we must take 2 of its three colors, so there are {11 \choose 1}{3 \choose 2} options. - z: There are now 10 options for the type that appears once, and for that type, we must take 1 of its three colors, so there are {10 \choose 1}{3 \choose 1} options. - So, the total number of options for this case is {12 \choose 1}{11 \choose 1}{3 \choose 2}{10 \choose 1}{3 \choose 1} = P(12, 3) 3^2 = {12 \choose 3} 3^3 \cdot 2.

So, the total number of options is {12 \choose 3} \cdot {3 \choose 2}^3 + {12 \choose 1}{11 \choose 1}{3 \choose 2}{10 \choose 1}{3 \choose 1}, which can be simplified in various ways, including to {12 \choose 3}3^4. Any of these ways received full credit, as long as they showed all work and logic.

\displaystyle\frac{{9 \choose 5} \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^4}{1 - \left( \frac{2}{3} \right)^9}. Anything equivalent to this answer received full credit.

This boils down to recognizing that P(\text{exactly 5 gold cards | at least one gold card}) = \frac{P(\text{exactly 5 gold cards})}{P(\text{at least one gold card})}, then breaking down the numerator and denominator.

The numerator, P(\text{exactly 5 gold cards}), is equal to {9 \choose 5} \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^4.

The denominator, P(\text{at least one gold card}), is equal to 1 - P(\text{no gold cards}) = 1 - \left( \frac{2}{3} \right)^9.

\displaystyle\frac{9!}{2!4!3!} (\frac{1}{12})^9. Anything equivalent to this answer received full credit, such as ${9 }{5 }{3 } ()^9 $ or {9 \choose 3}{6 \choose 4} (\frac{1}{12})^9.

On one individual draw, the probability of receiving one particular plane type is \frac{3}{36} = \frac{1}{12}, so the probability of seeing a particular sequence of 9 plane types – like A350, B767, A330, A330, B767, A350, A330, B767, B767 – is (\frac{1}{12})^9.

But, we need to account for the number of ways we can arrange 2 A350, 4 B767, and 3 A330 cards, which is \frac{9!}{2!4!3!} = {9 \choose 2} {7 \choose 4}{3 \choose 3} (which can also be expressed a few other ways, too).

\dfrac{{12 \choose 8}\cdot8!}{12^8} = \dfrac{12}{12}\cdot\dfrac{11}{12}\cdot\dfrac{10}{12}\cdot\dots\cdot\dfrac{5}{12}

There’s two ways you can think of this problem.

The numerator {12 \choose 8}\cdot8! is the number of total possible permutations of 8 distinct values; {12 \choose 8} possible sets of 8 distinct values times 8! orderings, because the die could roll the same set of distinct values in 8! different ways. The denominator 12^8 is the number of total possible outcomes. Altogether, dividing the numerator by the denominator gives us the probability of rolling 8 distinct values.

We can also solve this problem differently by realizing that the rolls are independent. In the first roll the probability of getting a new number is 1 (since any number would be new), now in the second roll the probability of getting a new number is \frac{11}{12} since we’ve only rolled one number so rolling any of the other 11 numbers would be rolling a new number.If you keep doing this for each roll, and multiply the independent probabilities of getting a new number in each of the rolls you’ll get \dfrac{12}{12}\cdot\dfrac{11}{12}\cdot\dfrac{10}{12}\cdot\dots\cdot\dfrac{5}{12}

\dfrac{{8 \choose 2}\cdot{12 \choose 7}\cdot7!}{12^8}

Again, I’m going to show you two different ways you can think of this problem. We can get the total number of outcomes that get you exactly 7 distinct values by doing the following, first choose 7 from the 12 possible numbers (there’s {12 \choose 7} ways to do it), then choose a number to duplicate (there’s {7 \choose 1} ways to do that), lastly you need to order them. From the 8 possible places you have availible you need to choose 2 for the repeated value (there’s {8 \choose 2} ways to do that), then once you chose a placement for your reapeted value you need to choose one for each of the remaing 6 values(there’s 6! ways to do that). Finally just multiply the different ways you can perform each of those actions and you’ll get the number of ways you can roll 7 different values \begin{align*} {12 \choose 7}\cdot{7 \choose 1}\cdot{8 \choose 2}\cdot6! = {8 \choose 2}\cdot{12 \choose 7}\cdot7! \end{align*}

lastly, just divide by the total number of possible outcomes and you get \dfrac{{8 \choose 2}\cdot{12 \choose 7}\cdot7!}{12^8}

You could also find the number of rolls with exactly 7 different values by doing the following. First, select an ordered string of size 7 from the 12 possible values (there’s {12 \choose 7}\cdot7! ways to do that), then take the first value of the string as the duplicated value and keep the others as they are, now you only need to place 2 copies of the repeated value somewhere on your string of 6 numbers (there’s {8 \choose 2} ways to do that), multiply this the ways you can do these two actions and divide by total number of outcomes and you’ll get \dfrac{{8 \choose 2}\cdot{12 \choose 7}\cdot7!}{12^8} This solution is faster but less intuitive.

{18 \choose 3} \cdot 3!

This problem is exactly asking for the number of permutations of three elements from a pool of 18 elements, which is \begin{align*} 18\cdot17\cdot16 = \frac{18!}{15!} = {18 \choose 3} \cdot 3! \end{align*}

You should remember that whenever you draw items without replacement and the order matters then you are calculating a permutation. Alternatively, you can reason that in the red segment of the plate, you can have any of the 18 different menu items, then for the yellow segment you only have 17 menu items available, since the red already has one, lastly you have 16 items left for the blue part of the plate. Then just multiply all the ways you can choose menu items for each segment getting a total of \begin{align*} 18\cdot17\cdot16 = {18 \choose 3} \cdot 3! \end{align*}

18^3

In this question, since we are allowed to repeat menu items, we have 18 options for each of the sections of the plate. Then when you multiply all the ways you can choose items for a secyion of the plate you get 18\cdot18\cdot18 = 18^3. Similarly to how we explained the question before this one, aske your self if the drawing process has replacement and if the order matters. In this case you’ll find that we have replacement and we also have ordeirng, so this is a sequence, and the number of sequence’s of lenght k from a pool of n elements is n^k.

2 \cdot {18 \choose 3}

Similarly to how we solved the previous problems, we are going to deconstruct the process of building a plate with the given characteristics. First, we know that any plate must have three items, so let’s start by choosing three items from the 18 menu items we have available (we have {18 \choose 3} ways to do that).

Now once you chose the menu items that go on your plate, you need to arange them on your plate, normally if we are aranging three distinct elements on a line we know that there’s 3! ways to do that, unfortunately like we saw in the prompt of this problem we need to account for when we rotate the plate. In the prompt it was signaled that the plate with Eggplant Tofu, Broccoli Beef, and fried rice in clockwise order around the plate, is the same as the plate with Broccoli Beef, fried rice, and Eggplant Tofu (also in clockwise order), this happens because if you rotate your fisrt plate in counter clockwise direction then you would get the second plate, and in fact we could do this again to show those two plates are the same as the plate with fried rice, Eggplant Tofu, and Broccoli Beef. If you were to do the rotation again you would get the original plate, so for any specific plate we can find three different lines that describe the same plate, thus if we divide the number of ways to order our 3 menu items on a line(3!) by 3, we should get the number of different plates we can build with any three items, so \frac{3!}{3} = 2

Finally we only need to mutiply the number of ways we can choose three menu items by the number of plates we can build with those three items getting 2 \cdot {18 \choose 3}.

2 \cdot {18 \choose 2} + {18 \choose 1}

Now that we are allowed to repeat menu items we need to consider the 2 new types of plates, type 1 a plate with the same menu item 3 times, type 2 a plate with a repeated menu item and an additional menu item. Then the total number of plates should be our answer from 4.3 plus the number of plates with type 1 and number of plates with type 2.

Type 1 Counting the number of type 1 plates should be fairly simple, if your plate only contains one menu item then there should be one plate of this type for each of the menu items, therefore there’s 18 plates like this.

Type 2 Now for type 2, we can start by chossing two items from the menu (there’s 18 \choose 2 ways to do that) then from here note that you need to choose which item you are going to repeat because depending on which one you choose to double you get a different plate (there’s 2 options to double), then you should also consider the order but you should realize that no matter how you arange the item on your plate you are just getting rotations of the same plate, therefore your total number of plates of type 2 is 2 \cdot {18 \choose 2}

Lastly, just add both the plates of type 1 and type 2, and that’s the number you need to add to your answer from 4.3. Answer: 2 \cdot {18 \choose 2} + {18 \choose 1}

12^n.

The key to solving this problem is the note “When there are multiple pieces of a given color and type (for example, 8 white pawns), we will assume they are indistinguishable from one another.” This means we count each type of chess piece for the colors black and white. There are 6 different types of chess pieces: pawn, bishop, knight, rook, queen, and king. There are 2 possible colors: black and white. This means there are 6 \cdot 2 = 12 unique colored types of chess pieces.

There are 12 options for the first person, there are 12 options for the second person, and so on, for n people, so we can write:

12 \cdot 12 \cdot \dots \cdot 12 = 12^n

We see there are 12^n possible results for this experiment because there are n people in this experiment who can each select 12 unique combinations.

\dfrac{8}{16}\cdot\dfrac{7}{15} = \dfrac{7}{30}

We get \dfrac{8}{16} from the fact there are 8 white pawns and 16 white chess pieces. So the chance of first getting a pawn is \dfrac{8}{16}.

Now we assume that we have a white pawn, meaning there are 8-1=7 white pawns left and 16 - 1 = 15 white chess pieces, which means we have a \dfrac{7}{15} chance for getting another white pawn.

From here we simplify the answer by writing:

\dfrac{8 \cdot 7}{16 \cdot 15} = \dfrac{56}{240} = \dfrac{7}{30}

Another way to solve this is by using combinatorics: Note that {8 \choose 2} means “8 choose 2.”

\dfrac{{8 \choose 2}}{16 \choose 2} = \dfrac{\frac{8}{32}\cdot\frac{7}{31}}{\frac{16}{32}\cdot\frac{15}{31}} = \dfrac{7}{30}

False.

We know that two probabilities are independent of each other if P(A \cap B) = P(A) \cdot P(B).

• Let P(A) be the probability of getting a white piece twice in a row.
• Let P(B) be the probability of getting a pawn twice in a row.

We can calculate the probability of getting a white piece twice in a row P(A) = \frac{16}{32} \cdot \frac{15}{31} = \frac{15}{62}. We get \frac{16}{32} from the fact there are 16 white pieces out of the 32 total pieces. We get \frac{15}{31} from the fact there are now 15 white pieces out of the 31 pieces left after taking out a white piece.

We can also calculate the probability of getting a pawn twice in a row. Note that the color of the pawn does not matter! P(B) = \frac{16}{32} \cdot \frac{15}{31} = \frac{15}{62}. We get \frac{16}{32} from the fact there are 16 pawns (8 black pawns plus 8 white pawns) out of the 32 total pieces. We get \frac{15}{31} from the fact there are now 15 pawns out of the 31 pieces left after taking out a pawn.

We can now calculate P(A) \cdot P(B): \frac{15}{62} \cdot \frac{15}{62} = \frac{225}{3844}

Recall we calculate P(A \cap B) = P(A) \cdot P(B|A).

We solved for P(B|A) in part A of the problem: \frac{7}{30}. We also already solved for P(A) above: \frac{15}{62}.

\frac{15}{62} \cdot \frac{7}{30} = \frac{7}{124}

We can see that P(A \cap B) \neq P(A) \cdot P(B) because \frac{7}{124} \neq \frac{225}{3844}, so the answer is “False.”

P(\text{long}) = \frac{7}{15}

We are given that: P(B_1)=P(B_2)=\frac{1}{2} P(\text{long}|B_1) = \frac{60}{60 + 40} = \frac{3}{5} P(\text{long}|B_2) = \frac{10}{10 + 20} = \frac{1}{3}

Using the law of total probability:

P(\text{long})=P(\text{long}|B_1) \cdot P(B_1)+P(\text{long}|B_2) \cdot P(B_2) = \frac{3}{5} \cdot \frac{1}{2} + \frac{1}{3} \cdot \frac{1}{2} = \frac{7}{15}

There are 3! \cdot 3! permutations.

We can treat the group of A, B, and C (in any order) as a single block of letters which we can call X. This will gurantee that any permutation we find will satisfy our condition of having A, B, and C being all together.

Our problem now has 3! = 6 permutations of our new letter bank: X, D, and E.

XDE XED DXE EXD DEX EDX

Let’s call these 6 permutations “structures”. For each of these structures, there are 3! = 6 rearrangements of the letters A, B, and C inside of X.

So, with 3! different arrangements of D, E and our block of letters (X), and with each of those arrangements individually having 3! ways to re-order the group of ABC, the overall number of permutations of ABCDE where ABC are all together, in any order, is 3! \cdot 3!.

There are 20 ways the series can be won.

The series ends when the last game in the series is the third win by either player. Overall, there must be at least 3 games to end a series (all wins by the same player) and at most 5 games to end a series (each player has two wins, and then a final win by either player) before either player wins 3 games. Let’s assume player A always wins, and then we’ll multiply the number of options by two to account for the possibilities where player B wins instead.

Breaking it down:

• For a series that ends in 3 games - there is a single option (all wins by player A).

• For a series that ends in 4 games, we’ve assumed the last game is a player A win to end the series, therefore the 3 preceding games have {3 \choose 2} options for player A winning 2 of the 3 preceding games.

• For a series that ends in 5 games, we’ve assumed the last game is a player A win to end the series, therefore the 4 preceding games have {4 \choose 2} options for player A winning 2 of the 4 preceding games.

Overall, the number of ways the series can occur is: 2 \cdot (1 + {3 \choose 2} + {4 \choose 2}) = 20

12^n

To answer this question we must determine how many different pieces each person can choose. In a complete chess set we know that there’s pawns, knights, bishops, rooks, queens, and kings. Then we also know that each of these pieces can be aither black or white, therefore there’s a totoal of 6 \cdot 2 = 12 different pieces. Notice that a complete chess set has 32 total pieces, but there’s multiple duplicates in those 32 pieces that is why we say there’s only 12 different pieces on a chess set.

Now since every person has their own chess set, we know that each of them can get any of the 12 pieces, therefore if we do this with one person there would be 12 possible outcomes, now if we do it with two persons the first person can get any of the 12 pieces, but then for each piece that person 1 can choose person 2 can choose from their 12 pieces, which means that in this case there’s 12 \cdot 12 outcomes. if we keep doing this for n persons we’ll get that there’s 12 \cdot \dots \cdot 12 = 12^n.

\frac{1}{4}

First, let’s calculate the probability of one person choosing a black pawn. We know that the complete chess set has 32 pieces, and out of those 32 there are 8 black pawns, which gives us a probability of \frac{8}{32} = \frac{1}{4} of selecting a black pawn. Now, think about how we interpret probabilities; for example, if we say that the probability of getting heads when flipping a coin is \frac{1}{2} that just means that if we flip a coin over and over, we expect to see heads \frac{1}{2} of the time. Similarly, when we say that the probability of choosing a black pawn is \frac{1}{4}, that means that if we were to choose a chess piece over and over, we expect to choose a black pawn \frac{1}{4} of the time.

As n becomes large, we choose chess pieces over and over n times. Therefore, you still expect to see the black pawn \frac{1}{4} of the time.

False.

We know that two probabilities are independent of each other if P(A \cap B) = P(A) \cdot P(B).

• Let P(A) be the probability of getting a white piece twice in a row.
• Let P(B) be the probability of getting a pawn twice in a row.

We can calculate the probability of getting a white piece twice in a row P(A) = \frac{16}{32} \cdot \frac{15}{31} = \frac{15}{62}. We get \frac{16}{32} from the fact there are 16 white pieces out of the 32 total pieces. We get \frac{15}{31} from the fact there are now 15 white pieces out of the 31 pieces left after taking out a white piece.

We can also calculate the probability of getting a pawn twice in a row. Note that the color of the pawn does not matter! P(B) = \frac{16}{32} \cdot \frac{15}{31} = \frac{15}{62}. We get \frac{16}{32} from the fact there are 16 pawns (8 black pawns plus 8 white pawns) out of the 32 total pieces. We get \frac{15}{31} from the fact there are now 15 pawns out of the 31 pieces left after taking out a pawn.

We can now calculate P(A) \cdot P(B): \frac{15}{62} \cdot \frac{15}{62} = \frac{225}{3844}

Recall we calculate P(A \cap B) = P(A) \cdot P(B|A).

To find P(B|A), we assume that both pieces are white and imagine the probabilities of getting two pawns from that batch of white pieces. To get the first pawn, the probability is \frac{8}{16} (8 pawns out of 16 white pieces). To get the second pawn, the probability is: \frac{7}{15} (7 remaining pawns out of 15 remaining white pieces). So, the probabilitiy of getting two pawns given we have two white pieces is P(B|A) = \frac{8}{16} \cdot \frac{7}{15} = \frac{7}{30}. We also already solved for P(A) above: \frac{15}{62}.

\frac{15}{62} \cdot \frac{7}{30} = \frac{7}{124}

We can see that P(A \cap B) \neq P(A) \cdot P(B) because \frac{7}{124} \neq \frac{225}{3844}, so the answer is “False.”

12^n

Each quark has 6 \cdot 2 = 12 possibilities because there are 6 possible types and 2 possible states. Since our experiment involves independently picking one of these 12 possibilities for each of our n quarks, the total number of results for the experiment is 12^n.

None of the above (72^n).

For making a meson, we need to choose two quarks. The first quark we choose has 6 \cdot 2 = 12 possibilities because there are 6 possible types and 2 possible states. However, our second quark cannot have the same state as our first; so the second quark has 6 \cdot 1 = 6 possibilities because there are still 6 possible types, but only 1 possible state.

This means a single meson has 12 \cdot 6 = 72 possibilities. (If the order of the quark and antiquark did not matter, it would instead have \frac{72}{2} = 36 possibilities). So, an experiment where we select n mesons uniformly at random has 72^n unique results.

{54 \choose 4} = 316251

Since order does not matter, we would use a combination instead of a permutation here. From 54 unique cards in our poker deck, we can choose 4 at random to get a card hand.

An alternative way to think about it: there are 54 options for the first card we pick, 53 for the second card, 52 for the third card, etc. For four cards, we have 54 \cdot 53 \cdot 52 \cdot 51 hands. However this will count the same hand in different card orders as different hands, which we don’t want. To correct this, we divide by 4! (4 \cdot 3 \cdot 2 \cdot 1), because each unique hand will appear 24 times but in different orders.

\frac{54 \cdot 53 \cdot 52 \cdot 51}{4 \cdot 3 \cdot 2 \cdot 1} = {54 \choose 4} = 316251

650

Since the first 4 cards are four-of-a-kind (same number), there are 13 ways to select a four-of-a-kind. For the 5th card, there are 50 total choices (12 remaining values \cdot 4 suits + 2 wildcards). So there are a total 13 \cdot 50 = 650 options.

P(\text{Four of a kind | Bomb}) = \frac{1}{103}

The number of hands containing two jokers (the other two cards are arbitrary) is given by: \begin{align*} \underbrace{{2 \choose 2}}_{\text{Two Jokers}} \cdot \underbrace{{52 \choose 2}}_{\text{Other Two Cards}} = \frac{52 \cdot 51}{2} = 1326. \end{align*} The number of hands that form four of a kind is 13, one for each of the 13 unique values in a deck.

For a bomb to occur, the four cards either contain two jokers or form four of a kind. Hence, \begin{align*} P(\text{Four of a kind | Bomb}) = \frac{13}{13 + 1326} = \frac{1}{103} \end{align*}

{36 \choose 6} = \frac{36!}{6!30!} = \frac{36 \cdot 35 \cdot 34 \cdot 33 \cdot 32 \cdot 31}{6!} = \frac{P(36, 6)}{6!}. Any of these answers, or anything equivalent to them, are correct.

There are 36 cards, and we must choose of 6 of them at random without replacement such that order matters.

{12 \choose 6} \cdot 3^6. Anything equivalent to this answer is also correct.

Since we’re asking for the number of hands with unique types, we first pick 6 different types, which can be done {12 \choose 6} ways. Then, for each chosen type, there are {3 \choose 1} = 3 ways of selecting the color. So, the colors can be chosen in 3 \cdot 3 \cdot ... \cdot 3 = 3^6 ways.

{12 \choose 2} = 66. Anything equivalent to this answer is also correct.

Once we pick 2 unique plane types, we need to choose 3 different colors per type, out of a set of 3 possibilities. This can be done, for each type, in {3 \choose 3} = 1 way.

{12 \choose 3} \cdot {3 \choose 2}^3 = {12 \choose 3} 3^3. Anything equivalent to this answer is also correct.

First, we choose 3 distinct types, which can be done in ${12 \choose 3} ways. Then, for each type, we choose 2 colors, which can be done in {3 \choose 2} ways per type.

{12 \choose 3} \cdot {3 \choose 2}^3 + {12 \choose 1}{11 \choose 1}{3 \choose 2}{10 \choose 1}{3 \choose 1}. Anything equivalent to this answer is also correct.

If we select 6 cards and end up with 3 unique plane types, there are two possibilities: - We get 2 of each type (e.g. xx yy zz). This is the case in part (d), so the total number of sets of 6 cards that fall here are {12 \choose 3} \cdot {3 \choose 2}^3. - We get 3 of one type, 2 of another, and 1 of the last one (e.g. xxx yy z). - xxx: There are 12 options for the type that appears 3 times, so that can be selected in {12 \choose 1} ways. We must take all 3 of that types colors, which can be done in {3 \choose 3} = 1 way. - yy: There are now 11 options for the type that appears 2 times, and for that type, we must take 2 of its three colors, so there are {11 \choose 1}{3 \choose 2} options. - z: There are now 10 options for the type that appears once, and for that type, we must take 1 of its three colors, so there are {10 \choose 1}{3 \choose 1} options. - So, the total number of options for this case is {12 \choose 1}{11 \choose 1}{3 \choose 2}{10 \choose 1}{3 \choose 1} = P(12, 3) 3^2 = {12 \choose 3} 3^3 \cdot 2.

So, the total number of options is {12 \choose 3} \cdot {3 \choose 2}^3 + {12 \choose 1}{11 \choose 1}{3 \choose 2}{10 \choose 1}{3 \choose 1}, which can be simplified in various ways, including to {12 \choose 3}3^4. Any of these ways received full credit, as long as they showed all work and logic.

\displaystyle\frac{{9 \choose 5} \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^4}{1 - \left( \frac{2}{3} \right)^9}. Anything equivalent to this answer received full credit.

This boils down to recognizing that P(\text{exactly 5 gold cards | at least one gold card}) = \frac{P(\text{exactly 5 gold cards})}{P(\text{at least one gold card})}, then breaking down the numerator and denominator.

The numerator, P(\text{exactly 5 gold cards}), is equal to {9 \choose 5} \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^4.

The denominator, P(\text{at least one gold card}), is equal to 1 - P(\text{no gold cards}) = 1 - \left( \frac{2}{3} \right)^9.

\displaystyle\frac{9!}{2!4!3!} (\frac{1}{12})^9. Anything equivalent to this answer received full credit, such as ${9 }{5 }{3 } ()^9 $ or {9 \choose 3}{6 \choose 4} (\frac{1}{12})^9.

On one individual draw, the probability of receiving one particular plane type is \frac{3}{36} = \frac{1}{12}, so the probability of seeing a particular sequence of 9 plane types – like A350, B767, A330, A330, B767, A350, A330, B767, B767 – is (\frac{1}{12})^9.

But, we need to account for the number of ways we can arrange 2 A350, 4 B767, and 3 A330 cards, which is \frac{9!}{2!4!3!} = {9 \choose 2} {7 \choose 4}{3 \choose 3} (which can also be expressed a few other ways, too).