GRE Quantitative Reasoning - How to find the greatest or least number of combinations

There are 20 people eligible for town council, which has three elected members.

Quantity A

The number of possible combinations of council members, presuming no differentiation among office-holders.

Quantity B

The number of possible combinations of council members, given that the council has a president, vice president, and treasurer.

Quantity A is greater.

Quantity B is greater.

The quantities are equal.

The relationship cannot be determined from the information given.

Explanation

This is a matter of permutations and combinations. You could solve this using the appropriate formulas, but it is always the case that you can make more permutations than combinations for all groups of size greater than one because the order of selection matters; therefore, without doing the math, you know that B must be the answer.

Clark is in the market for new capes. If the cape store sells 48 unique types of capes, a purchase of how many capes will correspond to the minimum amount of potential combinations of capes?

Explanation

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

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

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

In the case of being even:

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

In the case of being odd:

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

When a value of drifts farther from these values, the number of potential combinations decreases to a minimum of .

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

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

Since is even:

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

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

Quantity A: The number of possible combinations when three choices are made from six options.

Quantity B: The number of possible permutations when three choices are made from four options.

Quantity B is greater.

Quantity A is greater.

The two quantities are equal.

The relationship cannot be determined.

Explanation

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

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

With selections made from potential options, the total number of possible permutations(order matters) is:

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

Quantity A:

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

Quantity B:

$P(4,3)=\frac{4!}{(4-3)!}=24$

Quantity B is greater.

Claus is taking his twin brother Lucas out for ice cream. Claus knows that his brother is indecisive and wants to spend as little time choosing ice cream as possible. Claus can choose how many scoops Lucas can make for a sundae, as long as Lucas gets at least four. If there are twelve ice cream options, how many scoops should Claus tell Lucas to get?

Each scoop of ice cream is a unique flavor.

Explanation

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

With selections made from potential options, the total number of possible combinations is

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

In terms of finding the maximum number of combinations, the value of should be

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

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 from the max, the lower the amount of choices.

For this problem:

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

For the choices provided the greater difference from the max occurs for .

What is the minimum amount of handshakes that can occur among fifteen people in a meeting, if each person only shakes each other person's hand once?

210

105

32,760

250

Explanation

This is a combination problem of the form “15 choose 2” because the sets of handshakes do not matter in order. (That is, “A shakes B’s hand” is the same as “B shakes A’s hand.”) Using the standard formula we get: 15!/((15 – 2)! * 2!) = 15!/(13! * 2!) = (15 * 14)/2 = 15 * 7 = 105.

Mohammed is being treated to ice cream for his birthday, and he's allowed to build a three-scoop sundae from any of the thirty-one available flavors, with the only condition being that each of these flavors be unique. He's also allowed to pick different toppings of the available , although he's already decided well in advance that one of them is going to be peanut butter cup pieces.

Knowing these details, how many sundae combinations are available?

Explanation

Because order is not important in this problem (i.e. chocolate chip, pecan, butterscotch is no different than pecan, butterscotch, chocolate chip), it is dealing with combinations rather than permutations.

The formula for a combination is given as:

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

where is the number of options and is the size of the combination.

For the ice cream choices, there are thirty-one options to build a three-scoop sundae. So, the number of ice cream combinations is given as:

$C(31,3)=\frac{31!}{(31-3)!3!}=4495$

Now, for the topping combinations, we are told there are ten options and that Mohammed is allowed to pick two items; however, we are also told that Mohammed has already chosen one, so this leaves nine options with one item being selected:

$C(9,1)=\frac{9!}{(9-1)!1!}=9$

So there are 9 "combinations" (using the word a bit loosely) available for the toppings. This is perhaps intuitive, but it's worth doing the math.

Now, to find the total sundae combinations—ice cream and toppings both—we multiply these two totals:

There are 300 people at a networking meeting. How many different handshakes are possible among this group?

None of the other answers

45,000

89,700

44,850

300!

Explanation

Since the order of persons shaking hands does not matter, this is a case of computing combinations. (i.e. It is the same thing for person 1 to shake hands with person 2 as it is for person 2 to shake hands with person 1.)

According to our combinations formula, we have:

300! / ((300-2)! * 2!) = 300! / (298! * 2) = 300 * 299 / 2 = 150 * 299 = 44,850 different handshakes

Sammy is at an ice cream shoppe, aiming to build a sundae from two different flavors from a choice of thirty-one, and three separate toppings from a choice of ten. How many kinds of sundaes can he make?

Explanation

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

With selections made from potential options, the total number of possible combinations is

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

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:

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

Toppings:

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

The total number of potential sundaes is

If there are students in a class and people are randomly choosen to become class representatives, how many different ways can the representatives be chosen?

Explanation

To solve this problem, we must understand the concept of combination/permutations. A combination is used when the order doesn't matter while a permutation is used when order matters. In this problem, the two class representatives are randomly chosen, therefore it doesn't matter what order the representative is chosen in, the end result is the same. The general formula for combinations is $C(n,k)=\frac{n!}{(n-k)!k!}$ , where is the number of things you have and is the things you want to combine.