Each time the coin will show either heads or tails. Miranda can thus get $2\cdot2\cdot2\cdot2\cdot2 = 2^5 =\mathbf{32}$ different outcomes.

We can select each letter in $6$ different ways so we can create a total of $6^3 =\mathbf{216}$ different words.

We can only choose between $2$ different alternatives to the first letter: either $a$ or $e$. Because we are not allowed to choose the same letter twice, we have five choices left for the second letter. Finally, we have four choices left for the third letter. In total, we can create $2\cdot5\cdot4 =\mathbf{40}$ different words.

Gustav has ten choices for the flavour of his ice cream, two different choices on how the ice cream should be served and four different choices for topping (he can also choose not to have any topping at all!). Gustav can thus buy $10 ​​\cdot2\cdot4 =\mathbf{80}$ different ice creams.

This should be simple to you by now. Five different objects placed in a row can be arranged in $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = \mathbf{120}$ unique ways. 

The first student can choose between 20 different seats, the second student between 19 different seats, and so on until the fifteenth student who can choose between 6 different seats. There are thus $20 \cdot 19 \ldots 7 \cdot 6 = \mathbf{\dfrac{20!}{5!}}$ unique ways in which the students can be seated. Notice how we were able to rewrite $$20 \cdot 19 \ldots 7 \cdot 6 \ = \ \frac{20!}{5!}$$ By dividing with $5!$, we remove the last $5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ that we don't want in the multiplication. Keep this technique in mind. You'll see it many times again. You will also notice that we chose to present the final answer in the form $\dfrac{20!}{5!}$ instead of as a number. The reason for this is that factorials become very large very quickly. The number of ways the students can be seated is in fact so great that this number contains 17 digits! Because factorials get so large so quickly, we often prefer to keep them in factorial form if the answer is larger than 6 or 7 digits.

There are $3! = 3 \cdot 2 \cdot 1 = 6$ different ways to place the plants on the upper shelf. For each of these 6 permutations, there are $5 = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$ different ways to place the five pictures on the lower shelf. This means that there are $6 \cdot 120 = \mathbf{720}$ different ways to place all eight items on the shelves. 

$101!$ is the same as $101\cdot100\cdot (99!)$. $101!$ is therefore $101\cdot100 =\mathbf{10 \ 100}$ times larger than $99!$.

Six letters can be arranged in $6!$ permutations. However, the order of the two B:s and the two O:s don't matter. We therefore need to divide by $2!$ for each of them. The letters in BABOON can thus be rearranged to form $\dfrac{6!}{2! \ \cdot \ 2!} = \mathbf{180}$ words.  

a) Six objects can be arranged in $6!$ ways. However, because the pennies are identical it doesn't matter in which order we place them in so we need to divide by $2!$. Neither does it matter in which order we place the four nickels in, so we also need to divide by $4!$. This means that there are $\frac{6!}{2! \ \cdot \ 4!}=\mathbf{15}$ unique permutations of the six coins. 

b) Six objects can be arranged in $6!$ ways. But since we have two identical pennies, four identical dimes and three identical quarters, we need to divide by $2!$, by $4!$ and by $3!$ to avoid over-counting. This means that there are $\frac{9!}{2! \ \cdot \ 4! \ \cdot \ 3!}=\mathbf{1260}$ unique permutations of the nine coins.

Sara has 9 options when choosing her first book, 8 options for her second book and 7 options for her third book. 

If we multiply 9, 8 and 7, the product $9\cdot8\cdot7 = 504$ will include every possible way in which Sara can select three of her nine books. One of these $504$ ways is that she first chooses book $A$, then book $B$ and lastly book $C$. Another way is that she first chooses book $C$, then book $A$ and lastly book $B$. As you can see, we have a problem here. $\{A, B, C\}$ and $\{C, A, B\}$ are not two unique sets of books. They are two different permutations of the same set. How can we avoid counting the same set more than once? 

The same set of three books can be chosen in $3! = 6$ different ways. For example, if I chose the unique set $\{A, B, C\}$, I have three options for the first book, two options for the second and one option for the third book which is $3 \cdot 2 \cdot 1 = 3!$. Among the total $9\cdot8\cdot7 = 504$ ways to choose three books, there will thus be $3!=6$ copies of each unique book set. The number unique books sets that Sara can choose is therefore $\frac{9 \ \cdot \ 8 \ \cdot \ 7}{3!} =\mathbf{84}$.

This principle applies to any number of objects we choose. Had Sara chosen five books instead of three, she could have picked $\frac{9 \ \cdot \ 8 \ \cdot \ 7 \ \cdot \ 6 \ \cdot \ 5}{5!} =126$ unique book sets. 

Each of the ten neighbours shakes hands with nine other people. That neighbour $A$ shakes hands with neighbour $B$ is the same as that $B$ shakes hands with $A$ so in order not to overestimate we have to divide the number of greetings by 2. In total, there will be $\frac{10 \ \cdot \ 9}{2} =\mathbf{45}$ handshakes.

This question can actually be rephrased to "How many pairs of two people could we choose from a group of ten people?". Every pair will be equal to one handshake. Phrased this way, it's obvious that the answer is $\frac{10 \ \cdot \ 9}{2} =45$.

Sara could have chosen $\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3}{5!}=21$ sets of T-shirts, $\frac{6 \cdot 5 \cdot 4 \cdot 3}{4!}=15$ sets of skirts and $\frac{3 \cdot 2}{2!}=3$ sets of sneakers. Sara could therefore have chosen a total of $21 \cdot 15 \cdot 3=\mathbf{945}$ unique sets of clothes. 

a) We can select 20 objects from a set of 100 different objects in

$$\frac{100 \ \cdot \ 99 \ \ldots \ \cdot \ 82 \ \cdot \ 81}{20!}$$

different ways. However, we can express this in more simplified way. The numerator $100 \ \cdot \ 99 \ \cdot \ldots \ \cdot \ 82 \ \cdot \ 81$ can be rewritten as $\frac{100!}{80!}$. By dividing by $80!$, we remove the last 80 factors that we don't want in the product. We thus get:

$$\dfrac{100 \ \cdot \ 99 \ \cdot \ \ldots \ \cdot \ 82 \ \cdot \ 81}{20!} \ \ \ = \ \ \frac{\dfrac{100!}{80!}}{20!} \ \ = \ \ \mathbf{\frac{100!}{80! \ \cdot \ 20!}}$$

b) We can select $m$ objects from a set of $n$ different objects in

$$\frac{n \ \cdot \ (n - 1) \ \cdot \ \ldots \ \cdot \ (n - m + 2) \ \cdot \ (n - m + 1)}{m!}$$

different ways. Using the same principle as in question a), we can simplify this to:

$$\frac{n \ \cdot \ (n-1) \ \cdot \ \ldots \ \cdot \ (n-m+2) \ \cdot \ (n-m+1)}{m!} \ \ \ \ = \ \ \frac{\dfrac{n!}{(n-m)!}}{m!} \ \ = \ \ \dfrac{n!}{(n - m)! \ \cdot\ m!}$$

This expression is so common in combinatorics that it has been given its own special expression:

$$\dfrac{n!}{(n - m)! \ \cdot\ m!} \ = \ \binom{n}{m}$$

The symbol $\displaystyle{\binom{n}{m}}$ is pronounced as "$n$ choose $m$" because it represents the number of ways that $m$ objects can be selected from a set of $n$ different objects. Sometimes you will also see this expressed as $C_{m}(n)$, but $\displaystyle{\binom{n}{m}}$ is the most common of the two notations. The answer to question a) could thus also have been written as: 

$$\binom{100}{20} \ = \ C_{20}(100)$$

a) The teacher can select $\mathbf{\displaystyle{\binom{25}{3}}}$ sets of three students as class representatives. 

b) Because Gabriel has already chosen one of his four pinchos, he is actually only choosing three pinchos from a selection of 17 pinchos. Gabriel can therefore choose among $\mathbf{\displaystyle{\binom{17}{3}}}$ servings. 

c) In this scenario, the order in which the three best participants arrive matters. There are thus
$$\mathbf{20 \cdot 19 \cdot 18 = \dfrac{20!}{17!}}$$
possible outcomes in the race. Because the order matters, we don't divide by $3!$ as we otherwise do. 

d) There are $\displaystyle{\binom{30}{2}}$ pairs of senior members that could be chosen. Once the senior members are chosen, there are only 28 villagers left from which to choose the four junior members. There are thus
$$\mathbf{\displaystyle{\binom{30}{2}} \ \cdot \ \displaystyle{\binom{28}{4}}}$$
unique councils that could be formed. 

e) Simon can choose $\displaystyle{\binom{5}{3}}$ sets of plants. He may then choose among six pots for the first plant, among five pots for the second plant and among four pots for the third plant. Because the same three plants and the same three pots can be combined in unique ways, the order in which the pots are chosen matters. Simon thus has
$$\mathbf{\binom{5}{3} \ \cdot \ (6 \cdot 5 \cdot 4) \ = \  \binom{5}{3} \ \cdot \ \frac{6!}{3!}}$$

possible combinations of plants and pots to choose from. 

My sister could have chosen $\displaystyle{\binom{6}{3}}=\dfrac{6 \cdot \ 5 \ \cdot 4}{3!}=\mathbf{20}$ different ice creams. 

No matter which ice cream my sister chooses, there is going to be exactly one ice cream that doesn't have any similar flavours. For example, if my sister chooses the ice creams with flavours $\{1,2,3\}$, I can pick any combination of flavours except $\{4,5,6\}$. This means that I will always be able to pick $20-1=19$ possible ice creams. My sister and I can therefore choose $20 \cdot 19 = \mathbf{380}$ ice creams together. 

Samuel can choose his outfit in $3 \cdot 4 = \mathbf{12}$ different ways. The number of hats he can choose doesn't change depending on which jacket he picks. Therefore, we can just multiply the number of jackets and hats to determine the total number of combinations.

In this problem, Samuel's choice of hat now impacts the number of jackets he has available. There are fewer ways that Samuel can choose an outfit that includes his blue hat compared to outfits that doesn't include his blue hat. We should therefore divide the problem into different cases. 

(1) The hat Samuel chooses is blue

This gives Samuel one choice for his hat. Since he now won't choose either the brown or the black jacket, he only has one jacket left to choose from. Samuel therefore has only $1\cdot1 = 1$ possible outfit in this case.

(2) The hat Samuel chooses is not blue

This gives Samuel three hats to choose from and he may now choose any of his three jackets. Samuel therefore has  $3 \cdot 3 = 9$ possible outfits in this case.

In total, Samuel can thus combine his hats and jackets in $1 + 9 =\mathbf{10}$ different ways. 

NOTE! When you divide a task into different cases, it is very important that your cases really cover all possible alternatives and that there also is no overlap between the cases. Otherwise, you will either miss counting some outcomes or count some outcomes more than once. This is sometimes called the MECE Principle: Mutually Exclusive and Collectively Exhaustive. Mutually exclusive means that two cases can't happen at the same time, so there is no overlap. Collectively exhaustive means that our together cases cover all possible outcomes. 

The hundreds digit must be either 1 or 2. The integers that has the hundreds digit as one of their two similar digits, however, can be chosen in a different number of ways compared to the integers that doesn't have the hundreds digit as one of the two similar digits. We must therefore split up the problem into the two different cases:

(1) The hundreds digit is one of the two similar digits
We have two choices for the hundred digit: 1 or 2. One of the two remaining digits must be the same as the hundreds digit, but we can choose if this digit should be either in the tens place or the units place. The last digit we can choose freely from the ten existing digits. This case thus gives us $2\cdot2\cdot10 = 40$ outcomes.

(2) The hundreds digit is not one of the two similar digits
Again, we have two choices for the hundreds digit: 1 or 2. The tens and units digits must be the same, but we have nine different choices for which digit it should be. (We can choose any of the ten digits except the digit we chose for the hundreds digit). This case thus gives us $2 \cdot 9 = 18$ outcomes.

In total, there are thus $40 + 18 =\mathbf{58}$ positive integers between 100 and 299 that have exactly two similar digits.

This task problem can be divided into the following three cases:

(1) Everyone have at least one bill  (1,1,1)
This can only be done in 1 way: all children have one bill each.

(2) One student has no bills  (0,1,2)
One of the other two students thus has two bills and the third student has one bill. Because everyone has a different number of bills, this could be done in $3! = 6$ different ways.

(3) Two students have no bills  (0,0,3)
In this case, one of the children has all three bills. We can choose which child gets the three bills in 3 different ways.

Hugo, Wilhelmina and Rachel can thus divide the three bills between themselves in $1 + 6 + 3 = \mathbf{10}$ different ways.

Every four-digit number can be sorted into one of the five categories. The sum of all five cases therefore has to equal the total number of four-digit integers. But this means that we can find the answer by instead determining case (1): Four-digit integers no even digits, and subtracting that from the total! 

$$(1) + (2) + (3) + (4) + (5) \ = \ \text{Total Four-Digit Integers}$$
$$\text{Total Four-Digit Integers} - (1) \ = \ (2) + (3) + (4) + (5)$$

Let's first determine the total. The first digit in a number can't be 0, so we have nine options for the first digit but ten options for the other three. So in total there are $9 \cdot 10 \cdot 10  \cdot 10 = 9000$ four-digit integers. Now let's determine the number of four-digit integers with no even digits. In this case we have five options to choose from (1,3,5,7,9) for all four digits in the number so there $5 \cdot 5 \cdot 5 \cdot 5 = 625$ four-digit integers with no even digits. The number of four-digit integers with at least one even digit is therefore $9000-625=\pmb{8375}$.

When a problem can be split up into different cases, we have two methods to choose from. We can use either the direct addition method and count the things we want. Or we can use the indirect subtraction method and count the things we don't want and subtract that from the total. Be prepared to use whichever method that will require the least amount of work. When a problem includes phrases such as "at least", "at most" or "no more than", this is a clue that we need to determine the sum of several different cases. In these scenarios, the subtraction method is almost always going to be the fastest. 

This problem can be divided into five different cases. Sofia can choose: 

(1) 0 green teas
(2) 1 green tea
(3) 2 green teas
(4) 3 green teas
(5) 4 green teas

We want to calculate the sum of cases (2), (3), (4) and (5). We can either use the addition method and calculate all four cases separately. Or we calculate case (1), that none of the teas are red, and subtract it from the total. The subtraction method obviously requires the least amount of work.

Sofia can choose $\displaystyle{\binom{12}{3}}=\dfrac{12 \ \cdot \ 11 \ \cdot \ 10}{3!} = 220$ unique combinations of four teas in total. If no tea may be green, she has eight teas left to choose from. This can be done in $\displaystyle{\binom{8}{3}}=\dfrac{8 \ \cdot \ 7 \ \cdot \ 6}{3!} = 56$ different ways. Sofia can thus choose $220 - 56 =\pmb{164}$ combinations of teas so that at least one of them is green.

a) For each student, Patrick can choose between five different colors. He can thus give out pencils in $5^3 =\pmb{125}$ different ways.

Had the five council members been seated in a row there would have been $5!=120$ different seating arrangements. But in a circle, the council members can rotate their positions five times while still being seated in the same permutation. This means that there will only be $\dfrac{120}{5}=\pmb{24}$ unique ways in which the council can be seated. 

This pattern holds true generally. If we have $n$ objects arranged in a circle we can rotate the circle $n$ times. The number of unique permutations is therefore $\dfrac{n!}{n} = (n-1)!$. 

Another way to think about this is to fixate one of the objects. Because that object has to be in a certain place we can no longer rotate the circle. Now we have $(n-1)!$ objects left to arrange, and this time it will be like we arrange them in a row so the number of permutations again is $(n-1)!$. 

Let's fixate the position of person $A$. The other four elders can then be seated in $4!=24$ different ways. However, it's now possible to "flip" or "mirror" the entire seating arrangement with respect to person $A$ and still have everyone seated to the same two people. Consider the example below.

a) Keychains are circles in a 3D world. They can be rotated and also flipped upside-down with the keys still being in the same order. This means that there is only $\dfrac{3!}{3 \ \cdot \ 2}=\pmb{1}$ way in which three keys can be placed. This may seem counter-intuitive at first. After all, we would expect there to be more than just one keychain when three keys at our disposal. But it make sense if you think about how many different "neighbours" that each key could have. No matter how we place the keys, $A$ will always be between $B$ and $C$, $B$ will always be between $A$ and $C$, and $C$ will always be between $A$ and $B$. Viewed this way, it's obvious that there exists only one unique keychain with three keys. 

b) The fourth key can be inserted in three places - between $A$ and $B$, between $A$ and $C$, or between $B$ and $C$ - so there are $\pmb{3}$ unique keychains with four keys. We could also have solved this in the traditional way by dividing $4!$ first by $4$ (discarding rotations) and secondly by $2$ (discarding mirrorings) which gives us the same answer $\dfrac{4!}{4 \ \cdot \ 2}=\pmb{3}$. 

The king could choose the five jewels for the gold ring in $\displaystyle{\binom{15}{5}}=\dfrac{14 \ \cdot \ 13 \ \cdot \ 12 \ \cdot \ 11 \  \cdot \ 10}{5!} = 364$ ways. Since he now has $9$ jewels left, he can choose the jewels for the silver ring in $\displaystyle{\binom{9}{4}}=\dfrac{9 \ \cdot \ 8 \ \cdot \ 7 \ \cdot \ 6}{4!} = 126$ ways. The goldsmith now has $5$ jewels left for the bronze ring so he can choose which three jewels to use for this ring in $\displaystyle{\binom{5}{3}}=\dfrac{5 \ \cdot \ 4 \ \cdot \ 3}{3!} = 10$ ways. Now we need to consider the number of ways in which the jewels could be fastened around the three rings. 

A ring is a circle in 3D. It can therefore be rotated and flipped around while sporting the same permutation. We therefore need to divide by the number of "row-permutations" by the number of jewels in each ring to avoid overcounting the rotations. We also need to divide by $2$ for each ring to avoid counting the same ring when its flipped over. The number of ways in which the jewels can be placed around the three circles are thus as follows:

The gold ring:  $\dfrac{5!}{5 \ \cdot \ 2} = 12$ ways
The silver ring:  $\dfrac{4!}{4 \ \cdot \ 2} = 3$ ways 
The bronze ring:  $\dfrac{3!}{3 \ \cdot \ 2} = 1$ way

The goldsmith can thus present his king with $364 \cdot 126 \cdot 10 \cdot 12 \cdot 3 \cdot 1 = \pmb{16 \ 511 \ 040}$ unique sets of rings. 

The 6 student council members call 6 students each, so in the first tier there are $6\cdot6=36$ students.

Each student in the first tier then calls on 6 other students each, so in the second tier there are $36\cdot6=216$ students.

Each student in the second tier then calls on 6 other students each, so in the third tier there are $216\cdot4=864$ students. 

Since no student have received more than one call there are in total $6+36+216+864=\mathbf{1122}$ students who know about the meeting.

We could choose $\displaystyle{\binom{8}{3}}=\dfrac{8 \ \cdot \ 7 \ \cdot \ 6}{3!} =\pmb{56}$ sets of three different cookies.

We have five options for the first letter. For the second letter there are only four options, as there is always one letter that must not be followed directly by another. For the same reason, we also have four options to choose from for the third, the fourth and the fifth letter. All in all, there are $5\cdot4\cdot4\cdot4\cdot4 =\pmb{1280}$ good words.

The customer can choose $displaystyle{\binom{6}{3}} = \dfrac{6 \ \cdot \ 5 \ \cdot \ 4}{3!} = 20$ combinations of flavours and $displaystyle{\binom{3}{2}} = \dfrac{3 \ \cdot \ 2}{2!} = 3$ combinations of toppings. There are thus possible $20 \cdot 3 =\pmb{60}$ combinations with three flavours and two toppings that could be chosen. 

Books $A$ and $B$ can be placed in four different ways: places (1,2), (2,3), (3,4) or (4,5). For each of these placements, we can also choose if the two books should come in order $(A,B)$ or $(B,A)$. The last three books can now be placed in $3! = 6$ different ways. In total, we can arrange the five books in $4 \cdot 2 \cdot 6 =\pmb{48}$ different ways.

Solution 1
Claudia must have at least one \$5 bill as she can get a maximum of $12$ different sums if she has only twelve \$10 bills. The bills can be combined to all sums divisible by $5$ beginning with $5,10,\dots$. Since Claudia can form $17$ different sums, the $17$th sum must be $17 \cdot 5 = 85$.

If all the $12$ bills had been \$5 bills, the highest attainable sum would be $5 \cdot 12 = 60$. For each \$10 bill that is swapped for a \$5 bill, the sum of all bills increases by five. To get from $60$ to $85$, Claudia therefore needs to swap $\frac{85-60}{5}=5$ number of \$5 bills for \$10 bills. 

The number of \$10 bills that Claudia has is therefore $\pmb{5}$. 

Solution 2
Claudia must have at least one \$5 bill as she can get a maximum of $12$ different sums if she has only twelve \$10 bills. Since she has at least one \$5 she can get any sum that is a multiple of $5$, that is, $5, 10, 15, 20$… etc, until she reaches her maximum limit. In order for her to achieve exactly $17$ different sums, the largest number she can create must be $17 \cdot 5 = 85$. Let's say that Claudia has $x$ \$5 bills and $y$ \$10 bills. This gives us:

$$ x + y = 12$$

$$5x + 10y = 85 $$

which has the solution $(x, y) = (7, 5)$. Claudia thus has $\pmb{5}$ \$10 bills.

Once we have decided the placements for carts P and Y, the remaining four carts can always be placed in $4! = 24$ ways. If we can determine the number of possible placements of P and Y, we only need to multiply that result by $24$ to get the total number of ways the six carts could be assembled.

If P is in the first place, there are five places left for cart Y. If P is in the second place, there are four places left for cart Y, etc. By continuing this procedure we conclude that carts P and Y can be arranged in $5 + 4 + 3 + 2 + 1 = 15$ different ways. Thus, there are $15 \cdot 24 =\pmb{360}$ possible ways to assemble all the six the railway carts.

This problem can be divided into the following six cases:

(1) Five-digit integers with 0 odd digits

(2) Five-digit integers with 1 odd digit

(3) Five-digit integers with 2 odd digits

(4) Five-digit integers with 3 odd digits

(5) Five-digit integers with 4 odd digits

(6) Five-digit integers with 5 odd digits

The number of five-digit integers with at least two odd digits is the sum of cases (3), (4), (5) and (6). This easiest way to determine this is through the subtraction method. We determine the sum of cases (1) and (2) and subtract it from the total number of five-digit integers. There are $4\cdot5\cdot5\cdot5\cdot5 = 2500$ five-digit numbers with no odd digits. (The first digit can't be $0$ so we only have four even digits to choose from for the first digit instead of five). We will now determine the number of five-digit integers with exactly one odd digit. Depending on whether the first digit is odd or even, the number of options will differ. We therefore need to divide this case into the following two sub-cases:

(2.1) The first digit is odd
There are five options for the first digit. There are also five options for each of the remaining four digits, which all must be even. This case thus gives us $5\cdot5\cdot5\cdot5\cdot5 = 5^5 = 3125$ possible integers.

(2.2) The first digit is not odd
There are four options for the first digit: $2,4,6$ or $8$. We have five digits to choose from for the odd digit, and the odd digit can also be placed in four different places. There are five options each for the remaining three digits, which all must be even. This case thus gives us $4\cdot (5 \cdot 4) \cdot 5 \cdot 5 \cdot 5 = 10 \ 000$ possible integers.

There are thus $3125 + 10,000 = 13 125$ five-digit numbers with exactly one odd number. The total number of five-digit integers is $9 \cdot 10 \cdot 10 \cdot 10 \cdot1 0 = 9 \cdot1 0^4 = 90 \ 000$. All in all, there are $90 \ 000 - 2500 - 13 \ 125 =\pmb{74 \ 375}$ five-digit numbers that have at least two odd digits.

Let the number of guests at the party be $n$. Each guest shakes hands with $(n-1)$ other guests. But the order of the handshake doesn't matter ($A$ shaking hands with $B$ is the same $B$ shaking hands with $A$) we need to divide by $2$. The total number of handshakes is then $\frac{n(n - 1)}{2}$. One of $n$ or $(n - 1)$ must be odd. For the number of handshakes $\frac{n(n - 1)}{2}$ to be odd, either $n$ or $(n - 1)$ must be divisible by $2$ only once. The greatest integer less than $50$ which is only divisible by $2$ once is $46$. The greatest number of guests possible is thus $47$ which yields a total of $\frac{47 \ \cdot \ 46}{2} =\pmb{1081}$ handshakes.

Let us first determine the number of codes that don't include any of the digits $2, 0, 1, 7$. This means that we have $6$ options $(3,4,5,6,8,9)$ for each of the four digits digits, so the number of codes that don't include any of the digits $2,0,1,7$ is $6^4 = 1296$. The number of codes with at least one of the digits $2, 0, 1, 7$ is the same as the total number of codes minus the number of codes that don't include any of the digits $2, 0, 1 , 7$. The total number of codes is $10^4 = 10 \ 000$ and we already know that the number of codes that no digits in common with Christian's code is $1296$. This means that there are $10,000 - 1296 =\pmb{8704}$ codes that include at least one of the digits $2,0,1,7$, so this is the greater number. 

First off, we focus only on the ways the females can be arranged in aspect to one another. This is equivalent to calculating the number of different ways to arrange four females in a circle. We fix one female, say $A$, and we can see that there are $3$ different pairs of neighbours $A$ can have: $(B,C), (C,D)$ and $(B,D)$. 

Now we notice that male $a$ only can be positioned in two ways, given a configurations of females. When we choose a place for the $a$-male, the other positions are automatically determined. Thus, there are $3 \cdot 2 = \pmb{6}$ unique ways for the guests to be seated. These seating arrangements are:

$$–AcBdCaDb– \ \ \ \ \ –AdBaCbDc– \ \ \ \ \ –AdBcDaCb–$$
$$–AcBaDbCd– \ \ \ \ \ –AdCaBcDb– \ \ \ \ \ –AbCdBcD–$$

a) For each toy we can choose between $3$ shapes, $4$ materials, $5$ sizes and $6$ colors. The total number of possible toys is thus $3 \cdot 4 \cdot 5 \cdot 6 = \pmb{360}$.

b) All toys can be sorted into one of the following five categories:

(1) Toys that differ from a black, XL, plastic cube in $\pmb{0}$ ways
(2) Toys that differ from a black, XL, plastic cube in $\pmb{1}$ way
(3) Toys that differ from a black, XL, plastic cube in $\pmb{2}$ ways
(4) Toys that differ from a black, XL, plastic cube in $\pmb{3}$ ways
(5) Toys that differ from a black, XL, plastic cube in $\pmb{4}$ ways

The number of toys that differ from a black, XL, plastic cube in at least two ways is the sum of cases (3), (4) and (5). This easiest way to determine this is through the subtraction method. We determine the sum of cases (1) and (2) and subtract it from the total number of toys. 

(1) Toys different from a black, XL, plastic cube in $\pmb{0}$ways
There is only $1$ toys that are different from a black, XL, plastic cube in $0$ ways: that toy itself.

(2) Toys other than a black, XL, plastic cube in $\pmb{1}$ manner.
A toy can be different in either shape, material, size or color. If the toy is different in shape, there are $2$ options to choose from (sphere or tetrahedron). If the toy is different in material, there are $3$ options to choose from (wood, rubber or china). If the toy is different in size, there are $4$ options to choose from (L, M, S or XS). Last but not least, if the toy is different in color, there are $5$ options to choose from (white, yellow, green, red or blue). So there are $2 + 3 + 4 + 5 = 14$ toys that are different from a black, XL, plastic cube in exactly $1$ way.

This means that there are $360 - 14 - 1 =\pmb{345}$ toys that are different from a black, XL, plastic cube in at least two ways.