Understanding counting methods - GMAT Quantitative

The president can be elected in 70 different ways. After a student is elected president, there are 69 students left to elect a vice president from. Similarly, there are then 68 students left for the spot of treasurer. So there are different arrangements.

This is a permutation of 26 objects (letters) taken 4 at a time. Here order matters, because for example, "abcd" is not the same code word as "bdca".

You must know the permutation formula! It is as follows:

, where n is the number of different objects taken r at a time.

Here we have

Note: This is equivalent to 26 * 25 * 24 * 23.

Multiple the number of routes for each piece of the trip:

The number of subsets in a set of size is . If , then the set has subsets.

Alternatively, each subset of this twelve-element set is essentially a sequence of 12 independent decisions, one per element - each decision has two possible outcomes, exclusion or inclusion. By the multiplication principle, this is 2 taken as a factor 12 times, or

This can be seen, without loss of generality, as choosing each officer in turn.

There are 30 ways of choosing the president; there are then 29 ways of choosing the vice-president, and 28 ways of choosing the secretary-treasurer. Then 3 Student Senate representatives are chosen from the remaining 27 students; this is a combination of 3 elements from 27 - that is, . By the multiplication principle, the number of possible selections of the officers is:

There are eight prime numbers between 1 and 20:

The number of ways to select three of them, without regard to order, is the number of combinations of three out of eight:

Only one subset of has four even numbers - that is the subset .

Forming a subset with three even numbers and one odd number can be restated as choosing which even number to leave out and which odd number to include. There are 4 choices each way, so the number of sets fitting this description is .

Therefore, the number of subsets with at least three even elements is .

The finite series is obtained from by increasing each term by 1; since is an alternating series, this results in adding to :

so

Since one of Mick's choices is already decided, he will choose three people from a set of eleven without regard to order. This is a combination of three from a set of eleven; the number of such combinations is:

Claude will choose five people from a set of ten - twelve minus the two he dislikes - without regard to order. This is a combination of five from a set of ten; the number of such combinations is:

There are 24 ways to choose an allowed letter (since two of the 26 are excluded) and 10 ways to choose an allowed digit. Since repeats are allowed, and four letters and three digits will be chosen, there will be

possible license plate numbers.

Each of the four letters can be selected twenty-four ways, since two letters are forbidden and since repeats are allowed. Similarly, each of the three digits can be selected eight ways for the same reason.

By the multiplication principle, the number of possible license plate numbers is

Three different letters are selected from a group of 26, order being important - this is a permutation of three elements out of twenty-six. These permutations number .

There are then four numerals, each chosen from a set of ten. Since repeats are allowed, the number of ways this can be done is

By the multiplication principle, there are

possible license plates.

Each character of the licence plate number can be chosen from a set of 36 (26 letters plus 10 digits), and any character can be chosen multiple times. By the multiplication principle, this would allow different licence numbers. But licence numbers that comprise only letters and those that comprise only digits are disallowed, so by similar math, this reduces the possibilities by and , respectively.

Therefore, there are allowed licence plate numbers.

She has 4 choices for the shirt, 2 choices for the skirt per shirt chosen, and 3 choices for the pair of shoes per combination of shirt and skirt:

4 x 2 x 3 = 24 possible outfits

Just before Fred draws the 2nd card, there are 51 cards in the deck, because he already took the 1st card out.

Then once he puts the 2nd card back in, and shuffles, there are still 51 cards in the deck; he did not put the 1st card back in.

So Fred has a 1 in 51 chance of drawing the same card again.

Factorial numbers such as are evaluated as follows;

So to evaluate the expression

(Start)

(Expand the factorials)

(Cancel the common factors)

(Simplify)

Since the four letters are different, they can be arranged in different ways.

This combination problem asks us the number of ways to arrange four letters, including two that are the same. Four different letters can be arranged in different ways, but since we have two of the same letter, , we have to divide by . In any combination problem, if we have a total of letters, then for every number of the same letters, we have .

180 is the correct answer. This problem asks for the number of possible permutations of these six letters, which include two letters that are repeated twice, and . In any combination problem, if we have a total of letters, then for every number of the same letters, we have . So, this problem's situation can be modeled by the expression . We take the factorial of because we have six letters to be arranged. We divide by because we have two instances of letters being repeated twice.