Clara has three meats from which to choose (the four available, minus the one she will not eat); she has five vegetables from which to choose (the six available minus the one to which she is allergic). She can do one of the following:

Case 1: She can choose one meat and three vegetables.

She can choose one of three meats. 

She can choose three of five vegetables; the number of ways to do so is

$\displaystyle C(5,3)$

$\displaystyle = \frac{5!}{(5-3)! 3!}$

$\displaystyle = \frac{5!}{2! 3!}$

$\displaystyle = \frac{1 \cdot 2 \cdot 3 \cdot 4\cdot 5 }{1 \cdot 2 \cdot 1 \cdot 2\cdot 3 }$

$\displaystyle = \frac{ 4 \cdot 5 }{ 1 \cdot 2 }$

$\displaystyle = \frac{20}{2}$

$\displaystyle = 10$

The number of possible pizzas with one meat and three vegetables is 

$\displaystyle 3 \cdot C(5,3) = 3 \cdot 10 = 30$

Case 2: She can choose two meats and two vegetables.

She can choose two of three meats. There are three ways to do this, since choosing two meats means leaving one out; there are three ways to choose which meat to leave out.

She can choose two of five vegetables; the number of ways to do so is

$\displaystyle C(5,2)$

$\displaystyle = \frac{5!}{(5-2)! 2!}$

$\displaystyle = \frac{5!}{3! 2!}$

$\displaystyle = \frac{1 \cdot 2 \cdot 3 \cdot 4\cdot 5 }{1 \cdot 2 \cdot 3 \cdot 1 \cdot 2 }$

$\displaystyle = \frac{ 4 \cdot 5 }{ 1 \cdot 2 }$

$\displaystyle = \frac{20}{2}$

$\displaystyle = 10$

The number of possible pizzas with two meats and two vegetables is 

$\displaystyle 3 \cdot C(5,2) = 3 \cdot 10 = 30$

The total number of combinations of toppings from which Clara can choose is 

$\displaystyle 30+30 = 60$

Melissa paid $575 rent for each month from March to December - ten months - for a total of 

$\displaystyle \$ 575 \times 10 = \$5, 750$.

Add to this her prorated February rent and her security deposit (she was not late with any rent payments, hence no penalties); Melissa paid

$\displaystyle \$( 5, 750 + 250 +300)= \$6, 300$

Henry's rent is $645 per month; multiply this by 12 to get yearly rent before the penalty.

$\displaystyle \$ 645 \times 12 =\$ 7,740$

Henry paid his November rent three days late, so multiply the daily penalty of $20 by 3 to get the total penalty.

$\displaystyle \$20 \times 3 = \$ 60$

Add the regular yearly rent and the November penalty:

$\displaystyle \$ 7,740 + \$60 = \$7, 800$

The increment that is added to each successive term to obtain the next term is doubled each time.

$\displaystyle 7 + 1 = 8$

$\displaystyle 8 + 2 = 10$

$\displaystyle 10 + 4 = 14$

$\displaystyle 14+ 8 = 22$

$\displaystyle 22+ 16 = 38$

$\displaystyle 38 + 32 = 70$, the number in the circle

$\displaystyle 70 + 64 = 134$, the number in the square, which is the correct choice.

Beginning with the third term, each term is the sum of the previous two:

$\displaystyle 1+3 = 4$

$\displaystyle 3+4 = 7$

$\displaystyle 4+ 7 = 11$

$\displaystyle 7+11 = 18$

$\displaystyle 11+18 = 29$

$\displaystyle 18 + 29 = 47$, the number in the circle;

$\displaystyle 2 9+ 47 = 76$, the number in the square.

The number that is added to each term to obtain the next one increases by 2 each time:

$\displaystyle 4+ 4 = 8$

$\displaystyle 8+6 = 14$

$\displaystyle 14 + 8 = 22$

$\displaystyle 22+10 = 32$

$\displaystyle 32 + 12 = 44$

If we let $\displaystyle X$ be the number in the circle, then to maintain the pattern,

$\displaystyle X + 2 = 4$

so

$\displaystyle X = 2$. This is the correct choice. 

Each term, beginning with the third, is the sum of the previous two:

$\displaystyle 4+ 11 = 15$

$\displaystyle 11+ 15 = 26$

$\displaystyle 15+26 = 41$

$\displaystyle 26 + 41 = 67$

$\displaystyle 41+ 67 = 108$

If $\displaystyle X$ is the term in the circle, then 

$\displaystyle X+ 4 = 11$

and

$\displaystyle X = 7$.

In an arithmetic sequence, each term is obtained by adding the same number to the previous term - the common difference. Since $\displaystyle 7 - 1 = 6$, 6 is the common difference. The next two terms are:

$\displaystyle 7+6 = 13$

$\displaystyle 13+6 = 19$

19 replaces the square and is the correct choice.

In a geometric sequence, each term is obtained by muliplying the previous term by the same number - the common ratio. 

Since $\displaystyle 7 \div 1 = 7$, 7 is the common ratio. The next two numbers in the sequence are 

$\displaystyle 7 \times 7 = 49$

$\displaystyle 49 \times 7 = 343$

343 replaces the square; of the four choices, 350 comes closest and is the correct response.

There are three ways to choose a meat topping - four minus the one Mike won't eat.

There are five vegetable toppings - six minus the one to which John is allergic - from which to choose three. This is

$\displaystyle C(5,3)$

$\displaystyle = \frac{5!}{(5-3)! 3!}$

$\displaystyle = \frac{5!}{2! 3!}$

$\displaystyle = \frac{1 \cdot 2 \cdot 3 \cdot 4\cdot 5 }{1 \cdot 2 \cdot 1 \cdot 2\cdot 3 }$

$\displaystyle = \frac{ 4 \cdot 5 }{ 1 \cdot 2 }$

$\displaystyle = \frac{20}{2}$

$\displaystyle = 10$

The number of ways to make a Special Pizza that leaves out anchovies and mushrooms is 

$\displaystyle 3 \cdot C(5,3) = 3 \cdot 10 = 30$