This problem is asking us in how many ways the four different presents can be "arranged," or in this case, given to different people. Therefore, the answer is \(\displaystyle 4!\):

\(\displaystyle 4!=4\cdot3\cdot2\cdot1=24\)

The first thing that we should do is break down the different group of same letters in Tennessee: We have one "T," four "E's," two "N's," and two "S's." We have a total of nine letters; four of these are the same, as are two more pairs. In any combination problem, if we have a total of \(\displaystyle K\) letters, then for every \(\displaystyle n\) number of the same letters, we have \(\displaystyle \frac{K!}{n!}\) . So, we can model this problem's situation with the expression \(\displaystyle \frac{9!}{4!2!2!}\):

\(\displaystyle \frac{9!}{4!2!2!}\)

\(\displaystyle \frac{9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{(4\cdot3\cdot2\cdot1)(2\cdot1)(2\cdot1)}\)

\(\displaystyle \frac{362,880}{24*2*2}\)

\(\displaystyle \frac{362,880}{96} = 3780\)

This question asks us to count the number of ways to arrange 3 of the same items and 2 different items. Let's assign letters to each of them: "\(\displaystyle B\)" for one of the bicycles, \(\displaystyle S_1\) for the first type of skateboard, and \(\displaystyle S_2\) for the second type of skateboard. In this notation, our set looks like this: \(\displaystyle \begin{Bmatrix} B\: B\: B\: S_1\: S_2 \end{Bmatrix}\)

We have a total of five individuals to receive five items, one item each. In any combination problem, if we have a total of \(\displaystyle K\) letters, then for every \(\displaystyle n\) number of the same letters, we have \(\displaystyle \frac{K!}{n!}\) . So, for this situation, the total is given by dividing the total number of ways to arrange the items 5! by 3!, since 3 of these items are the same:

\(\displaystyle \frac{5!}{3!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{3\cdot2\cdot1}=\frac{120}{6}=20\)

Here, the number of "slots" is different than the number of letters we can use. We must first determine in how many ways we can select the letters to be arranged in our three-letter combination. Because we are selecting some but not all out of a presented group of options and order is not important, we can use the formula \(\displaystyle _nC_r=\frac{n!}{(n-r)!\cdot r!}\), where \(\displaystyle n\) is the total number of things from which we are choosing and \(\displaystyle r\) is the number things we are selecting for each group. In this case, \(\displaystyle n=4\) and \(\displaystyle r=3\), so our equation will look like this:

\(\displaystyle _nC_r=\frac{4!}{(4-3)!\cdot 3!}=\frac{4!}{1!\cdot 3!}\)

\(\displaystyle \frac{4\cdot3\cdot2\cdot1}{(1)(3\cdot2\cdot1)}=\frac{24}{1\cdot6}=4\)

We have 36 characters to choose from, since there are 10 single-digit numbers (0-9) and 26 letters in the alphabet. The length of the code is two characters and nothing was said about repeated combinations not being allowed, so the correct answer is given by \(\displaystyle 36^{2}\), which is \(\displaystyle 1296\).

In this problem, we are selecting four letters from a group of five letters that contains a pair of duplicate letters: two \(\displaystyle A\)s. We can model this situation using the formula \(\displaystyle \frac{n!}{r_1!r_2! . . .}\), where \(\displaystyle n\) is the number of things we have to choose from and each value of \(\displaystyle r\) corresponds to the number of duplicate items in a given group. In this case, we have five letters to pick from, so \(\displaystyle n=5\), and two of them are duplicates, so \(\displaystyle r=2\). This makes our equation:

\(\displaystyle \frac{5!}{2!}=\frac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=\frac{120}{2}=60\)

For each of the five "slots" that we want to have for our password, we have ten different digits to choose from. The number of passwords is given by calculating \(\displaystyle 10*10*10*10*10\) or \(\displaystyle 10^{5}\) , which equals \(\displaystyle 100,000\).

A three-letter-long password can be created in \(\displaystyle 26^{3}\) ways, since there are 26 letters to choose from and the password is three letters long.

\(\displaystyle 26^3=26\cdot26\cdot26=17,576\)

We are asked to find the number of ways in which the keys can be used on the locks. There are four keys and four locks. This is just like looking for the number of permutations of four different letters. Therefore, it is given by 4!

Indeed, if we start with the first lock, we have four keys, and if we only use the right key last, then we have made four attempts on the first lock. Assuming that we leave the key in the first lock, we then have three keys remaining for the second lock. Similarly, if we pick the wrong key twice, then the third opens the lock, and so on and so forth. We can then multiply all these possibilities \(\displaystyle (4\cdot3\cdot2\cdot1)\) to obtain \(\displaystyle 24\).

If a small pizza is chosen, one topping out of \(\displaystyle N+2\) can be ordered for no additional charge.

If a medium pizza is ordered, up to two different toppings out of \(\displaystyle N+2\) can be ordered for no additional charge - this is

\(\displaystyle C \left ( N+2, 2\right ) = \frac{\left (N+1 \right ) \cdot \left (N+2 \right )}{1 \cdot 2} = \frac{N^{2}+3N+2}{2}\)

If a large pizza is ordered, up to two different toppings can be ordered for no additional charge - again, this numbers \(\displaystyle \frac{N^{2}+3N+2}{2}\).

Add these expressions:

\(\displaystyle N+2 + \frac{N^{2}+3N+2}{2} + \frac{N^{2}+3N+2}{2}\)

\(\displaystyle = N+2 + N^{2}+3N+2\)

\(\displaystyle =N^{2}+4N + 4\)