Counting Methods - GMAT Quantitative Reasoning

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Question

The Department of Motor Vehicles wants to make all of the state's license numbers conform to two rules:

Rule 1: The license number must comprise three letters followed by four numerals.

Rule 2: Numerals can be repeated, but letters cannot.

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

Answer

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 $10 \times 10 \times 10 \times 10 = 10 ^{4}$

By the multiplication principle, there are

$P (26, 3)\times 10 ^{4}$

possible license plates.

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Question

In how many ways can the 11th grade class elect a president, vice president, and treasurer from a class of 70 students?

Answer

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 $\dpi{100} \small 70\times 69\times 68=328,440$ different arrangements.

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Question

What is the number of possible 4 letter code words that can be made from the alphabet, when all 4 letters must be different?

Answer

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:

${n}P{r}=\frac{n!}{(n-r)!}$ , where n is the number of different objects taken r at a time.

Here we have ${26}P{4}=\frac{26!}{(26-4)!} = \frac{26!}{22!}$

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

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Question

There are 8 paths between places and and 5 paths betweeen places and . How many different routes are there between places and ?

Answer

Multiple the number of routes for each piece of the trip: $8 \times 5 = 40$

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Question

How many subsets does a set with 12 elements have?

Answer

The number of subsets in a set of size is $\small 2^{n}$ . If , then the set has $\small \small \small \small 2^{12} = 4,096$ 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

$\small \small \small \small 2^{12} = 4096$

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Question

How many ways can a president, a vice-president, a secretary-treasurer, and three Student Senate representatives be selected from a class of thirty people? You may assume these will be six different people.

Answer

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, $C_{3}^{27}$ . By the multiplication principle, the number of possible selections of the officers is:

$N = 30 \cdot 29\cdot 28\cdot C_{3}^{27}$

$N= 30 \cdot 29\cdot 28\cdot \frac{ 27!}{(27-3)!; 3!}$

$N= 30 \cdot 29\cdot 28\cdot \frac{ 27!}{24!3!}$

$N= \frac{30 \cdot 29\cdot 28\cdot 27 \cdot26\cdot25}{3\cdot 2\cdot 1}$

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Question

How many ways can you select three different prime numbers between 1 and 20?

Answer

There are eight prime numbers between 1 and 20:

$\left { 2,3,5,7,11,13,17,19\right }$

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

$C(8,3) = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{6\cdot 7\cdot 8}{1\cdot 2\cdot 3}= 56$

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Question

Define set $A = \left { 1,2,3,4,5,6,7,8\right }$ .

How many four-element subsets of include at least three even numbers?

Answer

Only one subset of has four even numbers - that is the subset $\left { 2,4,6,8\right }$ .

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 $4\times 4 = 16$ .

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

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Question

Which of the following statements is true?

Answer

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

so

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Question

We want to create a three-character-long password using only the twenty-six letters from the English alphabet. How many different passwords can be created?

Answer

A three-letter-long password can be created in $26^{3}$ ways, since there are 26 letters to choose from and the password is three letters long.

$26^3=26\cdot26\cdot26=17,576$

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Question

Twelve students are running for student council; each student will vote for four. Mick wants to vote for his sister Janine. How many ways can he cast his ballot so as to include Janine among his choices?

Answer

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:

$C (11,3) = \frac{11!}{(11-3)!3!}= \frac{11!}{8!3!} = \frac{11 \times 10 \times 9}{3 \times 2\times 1} = \frac{990}{6} = 165$

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Question

Twelve students are running for student council; each student will vote for five. Claude does not want to vote for Gary or Mitch, neither of whom he likes. How many ways can Claude fill in the ballot so that he does not vote for Gary?

Answer

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:

$C (10,5) = \frac{10!}{(10-5)!; 5!} = \frac{10!}{ 5!; 5!} = \frac{3,628,800 }{ 120 \times 120} = 252$

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Question

The Department of Motor Vehicles wants to make all of the state's license numbers conform to three rules:

Rule 1: The license number must comprise three letters (A-Z), followed by three numerals (0-9), followed by a letter.

Rule 2: Repeats are allowed.

Rule 3: None of the letters can be an "O" or an "I".

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

Answer

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

$24 \times 24 \times 24 \times 10\times 10\times 10 \times 24 = 24 ^{4} \times 10 ^{3}$

possible license plate numbers.

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Question

The Department of Motor Vehicles wants to make all of the state's license numbers conform to four rules:

Rule 1: The license number must comprise two letters, followed by three digits, followed by two letters.

Rule 2: The letters "I" and "O" may not be used.

Rule 3: The digits "0" and "1" may not be used.

Rule 4: Repeats are allowed.

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

Answer

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

$24 \times 24 \times 8 \times 8 \times 8 \times 24 \times 24 = 24^{4} \times 8^{3}$

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Question

The Department of Motor Vehicles wants to make all of the state's license numbers conform to three rules:

Rule 1: Each license number must comprise six characters, each of which must be a letter or a digit.

Rule 2: No license number can comprise six letters or six digits.

Rule 3: Repeats are allowed.

Which of the following expressions is equal to the number of possible license numbers that would conform to this rule?

Answer

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 $36 ^{6}$ 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 $26 ^{6}$ and $10 ^{6}$ , respectively.

Therefore, there are $36 ^{6} - 26 ^{6} - 10^{6}$ allowed licence plate numbers.

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Question

Jessica went shopping and bought herself shirts, skirts, and pairs of shoes. How many outfits can Jessica possibly create from what she just bought?

Answer

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

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Question

Fred shuffles a deck of 52 cards. He then draws 1 card and keeps it. Then he draws a 2nd card. Then he puts it back into the deck and shuffles again. Fred then draws a 3rd card and keeps it

What's the probability that the 2nd and 3rd cards are the same?

Answer

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.

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Question

Simplify the following expression

$\frac{8!}{3!2!}$

Answer

Factorial numbers such as are evaluated as follows; $8! = 8\times7\times6\times5\times4\times3\times2\times\1$

So to evaluate the expression

$\frac{8!}{3!2!}$ (Start)

$\frac{8\times7\times6\times5\times4\times3\times2\times1}{3\times2\times1\times2\times1}$ (Expand the factorials)

$\frac{8\times7\times6\times5\times4}{2\times1}$ (Cancel the common factors)

(Simplify)

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Question

How many ways are there to arrange the letters , , , and ?

Answer

Since the four letters are different, they can be arranged in different ways. $4!=4\cdot3\cdot2\cdot1=24$

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Question

In how many ways can the letters , , and be arranged to form a four-letter combination?

Answer

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 $\frac{K!}{n!}$ .

$\frac{4!}{2!}=\frac{4\cdot3\cdot2\cdot1}{2\cdot1}=12$

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