Permutation / Combination - GRE Quantitative Reasoning

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Question

A chamber of commerce board has seven total members, drawn from a pool of twenty candidates. There are two stages in the board's election process. First, a president, secretary, and treasurer are chosen. After that, four members are chosen to be “at large” without any specific title or district. How many possible boards could be chosen?

Answer

We have to consider two cases. First, the group containing the president, secretary, and treasurer represents a case of permutation. Since the order matters in such a group, we can select from our initial 20 candidates 20 * 19 * 18, or 6840 possible groupings.

Following that, the at large group constitutes a case of combinations in which the order does not matter. Since we have chosen 3 already for the first three slots, there will be 17 remaining people. The formula for choosing 4-person sets out of 17 candidates is represented by the combination formula of this form:

17! / ((17-4)! * 4!) = 17! / (13! * 4!) = (17 * 16 * 15 * 14) / (4 * 3 * 2) = 17 * 4 * 5 * 7 = 2380

Thus, we have 6840 and 2380 possible groupings. These can each be combined with each other, meaning that we have 6840 * 2380, or 16,279,200 potential boards.

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Question

How many ordered samples of 5 cards can be drawn from a deck of 52 cards without replacement?

Answer

The key points we need to remember are that order matters and that we are sampling without replacement. This then becomes a simple permutation problem. We have 52 cards to be chosen 5 at a time, so the answer is 52 * 51 * 50 * 49 * 48.

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Question

Quantitative Comparison

3 cards are chosen from a standard 52-card deck.

Quantity A: The number of ways to choose 3 cards with replacement

Quantity B: The number of ways to choose 3 cards without replacement

Answer

Quantity A says with replacement, so we have 52 ways to choose the first card, then we replace it so we again have 52 ways to choose the 2nd card, and similarly we have 52 ways to choose the 3rd card. Therefore we have 52 * 52 * 52 ways of choosing 3 cards with replacement.

Quantity B says without replacement, so we have 52 ways to choose the first card, but then we don't put that card back in the deck so we have 51 ways to choose the second card. Again we don't put that card back, leaving 50 ways to choose the third card. Thus there are 52 * 51 * 50 ways of choosing 3 cards without replacement.

Clearly Quantity A is greater. In general, there should always be more ways to choose something with replacement than there are without replacement just as we showed above. If you already knew this, you could have picked Quantity A without the math. Notice that either way you come to the answer here, there is NO reason to finish the computations all the way through. This will save you time on many quantitative comparison problems. For example, we know that 52 * 52 * 52 is larger than 52 * 51 * 50 without actually figuring out what the two expressions equal.

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Question

In how many ways can five different colored balls be arranged in a row?

Answer

We need to arrange the 5 balls in 5 positions: _ _ _ _ _. The first position can be taken by any of the 5 balls. Then there are 4 balls left to fill the second position, and so on. Therefore the number of arrangements is 5! = 5 * 4 * 3 * 2 * 1 = 120.

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Question

In how many ways can 6 differently colored rose bushes be planted in a row that only has room for 4 bushes?

Answer

There are 6 ways to choose the first rose bush, 5 ways to choose the second, 4 ways to choose the third, and 3 ways to choose the fourth. In total there are 6 * 5 * 4 * 3 = 360 ways to arrange the rose bushes.

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Question

Daisy wants to arrange four vases in a row outside of her garden. She has eight vases to choose from. How many vase arrangements can she make?

Answer

For this problem, since the order of the vases matters (red blue yellow is different than blue red yellow), we're dealing with permutations.

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

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

$P(8,4)=\frac{8!}{(8-4)!}=1680$

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Question

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?

Answer

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.

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Question

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.

Answer

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.

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Question

The President has to choose from six members of congress to serve on a committee of three possible members. How many different groups of three could he choose?

Answer

This is a combinations problem which means order does not matter. For his first choice the president can choose from 6, the second 5, and the third 4 so you may think the answer is 6 * 5 * 4, or 120; however this would be the answer if he were choosing an ordered set like vice president, secretary of state, and chief of staff. In this case order does not matter, so you must divide the 6 * 5 * 4 by 3 * 2 * 1, for the three seats he’s choosing. The answer is 120/6, or 20.

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Question

Joe has a set of 10 books that he hasn't yet read. If he takes 3 of them on vacation, how many possible sets of books can he take?

Answer

He can choose from 10, then 9, then 8 books, but because order does not matter we need to divide by 3 factorial

(10 * 9 * 8) ÷ (3 * 2 * 1) = 720/6 = 120

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Question

How many different license passwords can one make if said password must contain exactly 6 characters, two of which are distinct numbers, another of which must be an upper-case letter, and the remaining 3 can be any digit or letter (upper- or lower-case) such that there are no repetitions of any characters in the password?

Answer

Begin by considering the three "hard and fast conditions" - the digits and the one upper-case letter. For the first number, you will have 10 choices and for the second 9 (since you cannot repeat). For the captial letter, you have 26 choices. Thus far, your password has 10 * 9 * 26 possible combinations.

Now, given your remaining options, you have 8 digits, 25 upper-case letters, and 26 lower-case letters (i.e. 59 possible choices). Since you cannot repeat, you will thus have for your remaining choices 59, 58, and 57 possibilities.

Putting all of this together, you have: 10 * 9 * 26 59 58 * 57 or 456426360 choices.

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Question

A restaurant has a meal special that allows you to choose one of three salads, one of five sandwiches, and two of fifteen side dishes. How many possible combinations are there for the meal?

Answer

Although this is a permutation style problem, we have to be careful regarding the last portion (i.e. the side dishes). We know that our meal will have:

(3 possible salads) * (5 possible sandwiches) * (x possible combinations of side dishes).

We must ascertain how many side dishes can be selected. Now, it does not matter what order we put together the side dishes, so we have to use the combinations formula:

c(n,r) = (n!) / ((n-r)! * (r!))

Plugging in, we get: c(15,2) = 15! / (13! * 2!) = 15 * 14 / 2 = 15 * 7 = 105

Using this in the equation above, we get: 3 * 5 * 105 = 1575

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Question

In how many different orders can 8 players sit on the basketball bench?

Answer

Using the Fundamental Counting Principle, there would be 8 choices for the first player, 7 choices for the second player, 6 for the third, 5 for the fourth, and so on. Thus, 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 or 8! = 40, 320.

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Question

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

Answer

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

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Question

What is the number of possible 4-letter words that can be made from the 26 letters in the alphabet, where all 4 letters must be different?

Assume non-sensical words count, i.e. "dnts" would count as a 4-letter word for our purposes.

Answer

This is a permutation of 26 letters taken 4 at a time. To compute this we multiply 26 * 25 * 24 * 23 = 358,800.

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Question

10 people want to sit on a bench, but the bench only has 4 seats. How many arrangements are possible?

Answer

The first seat can be filled in 10 ways, the second in 9 ways, the third in 8 ways, and the fourth in 7 ways. So the number of arrangements = 10 * 9 * 8 * 7 = 5040.

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Question

There are 16 members of a club. 4 will be selected to leadership positions. How many combinations of leaders are possible?

Answer

With permutations and combinations, you have to know if the order people are selected matters or not. If not, like in this case, you must take the number of people and positions available: $\dpi{100} \small 16 \times 15 \times \times 14 \times 13$ and divide by number of spots open $\dpi{100} \small 4 \times 3 \times 2 \times 1 = 1820$

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Question

How many different committees of 3 people can be formed from a group of 7 people?

Answer

There are $\dpi{100} \small 7\times 6\times 5=210$ different permutations of 3 people from a group of seven (when order matters). There are $\dpi{100} \small 3\times 2=6$ possible ways to arrange 3 people. Thus when order doesn't matter, there can be $\dpi{100} \small \frac{210}{6}=35$ different committees formed.

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Question

A restaurant serves its steak entree cooked rare, medium, or well done. The customer has the choice of salad or soup, with one of two salads or one of 4 soups. The customer also chooses between one of three soft drinks as well as water or milk. How many unique variations are there to the entire steak dinner of

steak + soup/salad + drink?

Answer

The customer has 3 choices on meat, 6 choices on side, and 5 choices on drink. This gives a total of $\dpi{100} \small 3 \times 5\times 6=90$ choices for the meal.

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Question

From a group of 8 students, 3 are attending a meeting.

Quantity A: The number of different groups that could attend among the 8 students

Quantity B: 336

Answer

To solve this problem, you would need to utilize the combination formula, which is $\dpi{100} \small C=\frac{n!}{r!(n-r!)}$ .

$\dpi{100} \small C$ is the number of possibilities, $\dpi{100} \small n$ is the number of students, and $\dpi{100} \small r$ is the students attending the meeting. Thus, $\dpi{100} \small \frac{8!}{3!5!}=56$ .

336 would be the result of computing the permutation, which would be incorrect in this case.

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Permutation / Combination - GRE Quantitative Reasoning

How many ordered samples of 5 cards can be drawn from a deck of 52 cards without replacement?

The key points we need to remember are that order matters and that we are sampling without replacement. This then becomes a simple permutation problem. We have 52 cards to be chosen 5 at a time, so the answer is 52 * 51 * 50 * 49 * 48.

Quantitative Comparison 3 cards are chosen from a standard 52-card deck. Quantity A: The number of ways to choose 3 cards with replacement Quantity B: The number of ways to choose 3 cards without replacement

In how many ways can five different colored balls be arranged in a row?

We need to arrange the 5 balls in 5 positions: _ _ _ _ _. The first position can be taken by any of the 5 balls. Then there are 4 balls left to fill the second position, and so on. Therefore the number of arrangements is 5! = 5 * 4 * 3 * 2 * 1 = 120.

In how many ways can 6 differently colored rose bushes be planted in a row that only has room for 4 bushes?

There are 6 ways to choose the first rose bush, 5 ways to choose the second, 4 ways to choose the third, and 3 ways to choose the fourth. In total there are 6 * 5 * 4 * 3 = 360 ways to arrange the rose bushes.

Daisy wants to arrange four vases in a row outside of her garden. She has eight vases to choose from. How many vase arrangements can she make?

For this problem, since the order of the vases matters (red blue yellow is different than blue red yellow), we're dealing with permutations. With selections made from potential options, the total number of possible permutations(order matters) is:

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?

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.

The President has to choose from six members of congress to serve on a committee of three possible members. How many different groups of three could he choose?

Joe has a set of 10 books that he hasn't yet read. If he takes 3 of them on vacation, how many possible sets of books can he take?

He can choose from 10, then 9, then 8 books, but because order does not matter we need to divide by 3 factorial (10 * 9 * 8) ÷ (3 * 2 * 1) = 720/6 = 120

A restaurant has a meal special that allows you to choose one of three salads, one of five sandwiches, and two of fifteen side dishes. How many possible combinations are there for the meal?

In how many different orders can 8 players sit on the basketball bench?

Using the Fundamental Counting Principle, there would be 8 choices for the first player, 7 choices for the second player, 6 for the third, 5 for the fourth, and so on. Thus, 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 or 8! = 40, 320.

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

What is the number of possible 4-letter words that can be made from the 26 letters in the alphabet, where all 4 letters must be different? Assume non-sensical words count, i.e. "dnts" would count as a 4-letter word for our purposes.

This is a permutation of 26 letters taken 4 at a time. To compute this we multiply 26 * 25 * 24 * 23 = 358,800.

10 people want to sit on a bench, but the bench only has 4 seats. How many arrangements are possible?

The first seat can be filled in 10 ways, the second in 9 ways, the third in 8 ways, and the fourth in 7 ways. So the number of arrangements = 10 * 9 * 8 * 7 = 5040.

There are 16 members of a club. 4 will be selected to leadership positions. How many combinations of leaders are possible?

With permutations and combinations, you have to know if the order people are selected matters or not. If not, like in this case, you must take the number of people and positions available: and divide by number of spots open

How many different committees of 3 people can be formed from a group of 7 people?

There are different permutations of 3 people from a group of seven (when order matters). There are possible ways to arrange 3 people. Thus when order doesn't matter, there can be different committees formed.

The customer has 3 choices on meat, 6 choices on side, and 5 choices on drink. This gives a total of choices for the meal.

From a group of 8 students, 3 are attending a meeting. Quantity A: The number of different groups that could attend among the 8 students Quantity B: 336

To solve this problem, you would need to utilize the combination formula, which is . is the number of possibilities, is the number of students, and is the students attending the meeting. Thus, . 336 would be the result of computing the permutation, which would be incorrect in this case.

Quantitative Comparison

3 cards are chosen from a standard 52-card deck.

Quantity A: The number of ways to choose 3 cards with replacement

Quantity B: The number of ways to choose 3 cards without replacement

He can choose from 10, then 9, then 8 books, but because order does not matter we need to divide by 3 factorial

(10 * 9 * 8) ÷ (3 * 2 * 1) = 720/6 = 120

What is the number of possible 4-letter words that can be made from the 26 letters in the alphabet, where all 4 letters must be different?

Assume non-sensical words count, i.e. "dnts" would count as a 4-letter word for our purposes.

The customer has 3 choices on meat, 6 choices on side, and 5 choices on drink. This gives a total of $\dpi{100} \small 3 \times 5\times 6=90$ choices for the meal.

From a group of 8 students, 3 are attending a meeting.

Quantity A: The number of different groups that could attend among the 8 students

Quantity B: 336