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

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

Quantity A:

\(\displaystyle P(10,7)=\frac{10!}{(10-7)!}=604800\)

Quantity B:

\(\displaystyle P(11,5)=\frac{11!}{(11-5)!}=55440\)

Quantity A is greater.

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

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

Quantity A:

\(\displaystyle C(8,4)=\frac{8!}{(8-4)!4!}=70\)

Quantity B:

\(\displaystyle C(8,5)=\frac{8!}{(8-5)!5!}=56\)

Quantity A is greater.

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

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

We'll be dealing with the potential combinations for the scarves, socks, and sweaters; the total amount of combinations will be the product of these three.

Scarves:

\(\displaystyle C(12,2)=\frac{12!}{(12-2)!2!}=66\)

Socks:

\(\displaystyle C(10,4)=\frac{10!}{(10-4)!4!}=210\)

Sweaters:

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

The total number of combiations is

\(\displaystyle (66)(210)(20)=277200\)

He's certainly not hurting for choices.

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.