The sequence is formed by alternately adding \(\displaystyle 4\) and adding \(\displaystyle 5\) to each term to get the next term.

\(\displaystyle \begin{matrix} 10 + 4 = 14\\ 14 + 5 = 19\\ 19+4= 23\\ 23+5=28\\ 28+4 = 32\\ 32+5=37\\ 37+4=41\\ 41+5=46 \end{matrix}\)

\(\displaystyle 41\) and \(\displaystyle 46\) are the next two numbers.

We demonstrate that all of the choices are subsets of \(\displaystyle A \cap B\).

\(\displaystyle A \cap B\) is the intersection of \(\displaystyle A\) and \(\displaystyle B\) - that is, the set of all elements of both sets. Therefore, 

\(\displaystyle A \cap B = \left \{ c,e,f,h,j,p\right \}\)

\(\displaystyle \left \{ c,e,f,h,j,p\right \}\) itself is one of the choices; it is a subset of itself. The empty set \(\displaystyle \varnothing\) is a subset of every set. The other two sets listed comprise only elements from \(\displaystyle A \cap B\), making them subsets of \(\displaystyle A \cap B\).

We are looking for an element that is in the intersection of \(\displaystyle A\) and \(\displaystyle B\) - in other words, we are looking for an element that appears in both sets.

\(\displaystyle A\) is the set of all multiples of 8. We can eliminate two choices as not being in \(\displaystyle A\) by demonstrating that dividing each by 8 yields a remainder:

\(\displaystyle 484 \div 8 = 60 \textrm{ R }4\)

\(\displaystyle 900 \div 8 = 112 \textrm{ R }4\)

\(\displaystyle B\) is the set of all perfect square integers.  We can eliminate two additional choices as not being perfect squares by showing that each is between two consecutive perfect squares:

\(\displaystyle 18^{2}= 324< 352 < 361 = 19^{2}\)

\(\displaystyle 18^{2}= 324< 336 < 361 = 19^{2}\)

This eliminates 352 and 336. However, 

\(\displaystyle 256 = 16 ^{2}\).

It is also a multiple of 8:

\(\displaystyle 256 \div 8 = 32\)

Therefore, \(\displaystyle 256 \in A \cap B\).

\(\displaystyle A \cup B\) is the union of \(\displaystyle A\) and \(\displaystyle B\), the set of all elements that appear in either set. Therefore, we are looking to eliminate the elements in \(\displaystyle A\) and those in \(\displaystyle B\) to find the element in neither set.

\(\displaystyle A\) is the set of all multiples of 8. We can eliminate two choices as mulitples of 8:

\(\displaystyle 272 \div 8 = 34\), so \(\displaystyle 272 \in A\)

\(\displaystyle 344\div 8 = 43\), so \(\displaystyle 344 \in A\)

\(\displaystyle B\) is the set of all perfect square integers. We can eliminate two additional choices as perfect squares:

\(\displaystyle \sqrt{225}=15\), so \(\displaystyle 225 \in B\)

\(\displaystyle \sqrt{441}=21\), so \(\displaystyle 441 \in B\)

All four of the above are therefore elements of \(\displaystyle A \cup B\).

420, however is in neither set:

\(\displaystyle 420 \div 8 = 52 \textrm{ R } 4\), so \(\displaystyle 420 \notin A\)

and 

\(\displaystyle 20 = \sqrt{400}< \sqrt{420} < \sqrt{441} = 21\), so \(\displaystyle 420 \notin B\)

Therefore,  \(\displaystyle 420 \notin A \cup B\), making this the correct choice.

Derreck is choosing three students from a field of six (seven minus Anne) without respect to order, making this a combination. He has \(\displaystyle C(6,3)\) ways to choose. This is:

\(\displaystyle C(6,3)= \frac{6!}{(6-3)! \; 3! } = \frac{6!}{3! \; 3! } = \frac{720 }{6 \times 6}= 20\) 

Derreck has 20 ways to fill the ballot.

Four candidates are being selected from ten, with order being important; this means that we are looking for the number of permutations of four chosen from a set of ten. This is

\(\displaystyle P (10,4) = \frac{10!}{(10-4)!}= \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5,040\)

There are 5,040 ways to complete the ballot.

These are four independent events, so by the multiplication principle, the ballot can be filled out \(\displaystyle 4 \times 5 \times 4\times 7 = 560\) ways.

These are three independent events, so by the multiplication principle, the ballot can be filled out \(\displaystyle 6 \times 5 \times 6 = 180\) ways.

Since Mike is already chosen, Roy is in essence choosing four candidates from nine, with order being important. This is a permutation of four elements out of nine. The number of these is

\(\displaystyle P (9,4) = \frac{9!}{(9-4)!} = \frac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3,024\)

Roy can fill out the ballot 3,024 times and have Mike be his first choice.

To find the next number in the list, multiply the previous number by \(\displaystyle \small 3\).

\(\displaystyle 27\times3=81\)