For this problem, order matters! Wearing a particular blouse on Friday is not the same as wearing it on Sunday. So that means that this problem will be dealing with permutations.

With $\displaystyle k$ selections made from $\displaystyle n!$ potential options, the total number of possible permutations(order matters) is:

$\displaystyle P(n,k)=\frac{n!}{(n-k)!}$

What we'll do is calculate the number of permutations for her blouses, skirts, and shoes seperately (determining how the Friday/Saturday/Sunday blouses/skirts/shoes could be decided), and then multiply these values.

Blouses:

$\displaystyle P(10,3)=\frac{10!}{(10-3)!}=720$

Skirts:

$\displaystyle P(12,3)=\frac{12!}{(12-3)!}=1320$

Shoes:

$\displaystyle P(8,3)=\frac{8!}{(8-3)!}=336$

Thus the number of potential outfit assignments is

$\displaystyle 720(1320)(336)=319334400$

Since in this problem the order of selection does not matter, we're dealing with combinations.

With $\displaystyle k$ selections made from $\displaystyle n!$ potential options, the total number of possible combinations is

$\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}$

Sammy is making two sub combinations; one of ice cream and one of toppings. The total amount of combinations will be the product of these two.

Ice cream:

$\displaystyle C(31,2)=\frac{31!}{(31-2)!2!}=465$

Toppings:

$\displaystyle C(10,3)=\frac{10!}{(10-3)!3!}=120$

The total number of potential sundaes is

$\displaystyle (465)(120)=55800$

Since in this problem the order of selection does not matter, we're dealing with combinations.

With $\displaystyle k$ selections made from $\displaystyle n!$ potential options, the total number of possible combinations is

$\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}$

$\displaystyle C(45,13)=\frac{45!}{(45-13)!13!}=73006209045$

Jeez, Jessie, go easy.

Since in this problem the order of selection does not matter, we're dealing with combinations.

With $\displaystyle k$ selections made from $\displaystyle n!$ potential options, the total number of possible combinations is

$\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}$

In terms of finding the maximum number of combinations, the value of $\displaystyle k$ should be 

$\displaystyle k_{max}=\frac{n_{even}}{2};k_{max}=\frac{n_{odd}\pm 1}{2}$

Since there are twelve options, a selection of six scoops will give the maximum number of combinations.

$\displaystyle C(12,6)=\frac{12!}{(12-6)!6!}=924$

$\displaystyle C(12,7)=792$

$\displaystyle C(12,8)=495$

$\displaystyle C(12,9)=220$

$\displaystyle C(12,10)=66$

Since in this problem the order of selection does not matter, we're dealing with combinations.

With $\displaystyle k$ selections made from $\displaystyle n!$ potential options, the total number of possible combinations is

$\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}$

In terms of finding the maximum number of combinations, the value of $\displaystyle k$ should be 

$\displaystyle k_{max}=\frac{n_{even}}{2};k_{max}=\frac{n_{odd}\pm 1}{2}$

Once the number of choices goes above or below this value (or below the minimum kmax/above the maximum kmax for an odd number of max choices), the number of potential combinations decreases. The farther the value of $\displaystyle k$ from the max, the lower the amount of choices.

For this problem:

$\displaystyle k_{max}=\frac{12}{2}=6$

For the choices provided the greater difference from the max occurs for $\displaystyle k=9$.

$\displaystyle |9-6|=3$

$\displaystyle |8-6|=2$

$\displaystyle |7-6|=1$

$\displaystyle |6-6|=0$

$\displaystyle |5-6|=1$

Since in this problem the order of selection does not matter, we're dealing with combinations.

With $\displaystyle k$ selections made from $\displaystyle n!$ potential options, the total number of possible combinations is

$\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}$

In terms of finding the maximum number of combinations, the value of $\displaystyle k$ should be 

$\displaystyle k_{max}=\frac{n_{even}}{2};k_{max}=\frac{n_{odd}\pm 1}{2}$

Since there is an odd number of cars:

$\displaystyle k_{max}=\frac{23\pm 1}{2}=\frac{24}{2},\frac{22}{2}=11,12$

Of course, it is not possible to purchase half a set.

Since in this problem the order of selection does not matter, we're dealing with combinations.

With $\displaystyle k$ selections made from $\displaystyle n!$ potential options, the total number of possible combinations is

$\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}$

In terms of finding the maximum number of combinations, the value of $\displaystyle k$ should be 

$\displaystyle k_{max}=\frac{n_{even}}{2};k_{max}=\frac{n_{odd}\pm 1}{2}$

Once the number of choices goes above or below this value (or below the smaller k_max/above the greater k_max for an odd number of total options), the number of potential combinations decreases. The farther the value of $\displaystyle k$ from the max, the lower the amount of choices.

In other words:

$\displaystyle |k-k{max}|_{up}\rightarrow combinations_{down}$

We're given an odd number of options so,

$\displaystyle k_{max}=\frac{11 \pm 1}{2}=5,6$

For the available choices 3, 4, 5, 6, or 7:

$\displaystyle |3-5|=2$

$\displaystyle |4-5|=1$

$\displaystyle |5-5|=0$

$\displaystyle |6-6|=0$

$\displaystyle |7-6|=0$

$\displaystyle 3$ will give the minimum number of choices.

With $\displaystyle k$ selections made from $\displaystyle n!$ potential options, the total number of possible combinations (order doesn't matter) is:

$\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}$

The number of combinations increases the closer the value of $\displaystyle k$ is to $\displaystyle \frac{1}{2}n$.

In the case of $\displaystyle n$ being even:

$\displaystyle k_{max}=\frac{n_{even}}{2}$

In the case of $\displaystyle n$ being odd:

$\displaystyle k_{max_{1,2}}=\frac{n_{odd} \pm 1}{2}$

When a value of $\displaystyle k$ drifts farther from these values, the number of potential combinations decreases to a minimum of $\displaystyle n$.

$\displaystyle |k-k_{max}|_{up} \rightarrow #Combinations_{down}$

Note that for an odd $\displaystyle n$, consider the difference small values of $\displaystyle k$ $\displaystyle (< k_{max_{lower}})$ and the smaller $\displaystyle k_{max}$, and the difference of large values of $\displaystyle k$ $\displaystyle (>k_{max_{greater}})$ and the larger $\displaystyle k_{max}$.

Since there are 37 options, an odd number:

$\displaystyle k_{max_{1,2}}=\frac{37 \pm 1}{2}=18,19$

For the potential numbers of purchased sweaters:

$\displaystyle |16-18|=2$

$\displaystyle |17-18|=1$

$\displaystyle |18-18|=0$

Note that nineteen also corresponds to the maximum number of possible combinations.

$\displaystyle |19-19|=0$

$\displaystyle |20-19|=1$

$\displaystyle k=16$ gives the smallest amount of potential combinations for the choices presented.

With $\displaystyle k$ selections made from $\displaystyle n!$ potential options, the total number of possible combinations (order doesn't matter) is:

$\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}$

The number of combinations increases the closer the value of $\displaystyle k$ is to $\displaystyle \frac{1}{2}n$.

In the case of $\displaystyle n$ being even:

$\displaystyle k_{max}=\frac{n_{even}}{2}$

In the case of $\displaystyle n$ being odd:

$\displaystyle k_{max_{1,2}}=\frac{n_{odd} \pm 1}{2}$

When a value of $\displaystyle k$ drifts farther from these values, the number of potential combinations decreases to a minimum of $\displaystyle n$.

$\displaystyle |k-k_{max}|_{up} \rightarrow #Combinations_{down}$

Note that for an odd $\displaystyle n$, consider the difference small values of $\displaystyle k$ and the smaller $\displaystyle k_{max}$, and the difference of large values of $\displaystyle k$ and the larger $\displaystyle k_{max}$.

Since $\displaystyle n$ is even:

$\displaystyle k_{max}=\frac{48}{2}=24$

$\displaystyle k:24;|24-24|=0;C(48,24)=32247603683100$

$\displaystyle k:25;|25-24|=1;C(48,25)=30957699535776$

$\displaystyle k:26;|26-24|=2;C(48,26)=27385657281648$

$\displaystyle k:27;|27-24|=3;C(48,27)=22314239266528$

$\displaystyle k:28;|28-24|=4;C(48,28)=16735679449896$

$\displaystyle 28$ is farthest from $\displaystyle k_{max}$ and gives the least amount of possible combinations.

Since in this problem the order of selection does not matter, we're dealing with combinations.

With $\displaystyle k$ selections made from $\displaystyle n!$ potential options, the total number of possible combinations is

$\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}$

Quantity A:

$\displaystyle C(10,3)=\frac{10!}{(10-3)!3!}=120$

Quantity B:

$\displaystyle C(10,7)=\frac{10!}{(10-7)!7!}=120$

The two quantities are equal.