Permutation / Combination - GRE Quantitative Reasoning
Card 1 of 384
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?
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?
Tap to reveal 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.
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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How many ordered samples of 5 cards can be drawn from a deck of 52 cards without replacement?
How many ordered samples of 5 cards can be drawn from a deck of 52 cards without replacement?
Tap to reveal 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.
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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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
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
Tap to reveal 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.
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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In how many ways can five different colored balls be arranged in a row?
In how many ways can five different colored balls be arranged in a row?
Tap to reveal 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.
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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In how many ways can 6 differently colored rose bushes be planted in a row that only has room for 4 bushes?
In how many ways can 6 differently colored rose bushes be planted in a row that only has room for 4 bushes?
Tap to reveal 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.
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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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?
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?
Tap to reveal 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:
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
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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?
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?
Tap to reveal 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.
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.
← Didn't Know|Knew It →
How many ordered samples of 5 cards can be drawn from a deck of 52 cards without replacement?
How many ordered samples of 5 cards can be drawn from a deck of 52 cards without replacement?
Tap to reveal 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.
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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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
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
Tap to reveal 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.
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.
← Didn't Know|Knew It →
In how many ways can five different colored balls be arranged in a row?
In how many ways can five different colored balls be arranged in a row?
Tap to reveal 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.
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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In how many ways can 6 differently colored rose bushes be planted in a row that only has room for 4 bushes?
In how many ways can 6 differently colored rose bushes be planted in a row that only has room for 4 bushes?
Tap to reveal 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.
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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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?
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?
Tap to reveal 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:
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
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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?
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?
Tap to reveal 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.
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.
← Didn't Know|Knew It →
How many ordered samples of 5 cards can be drawn from a deck of 52 cards without replacement?
How many ordered samples of 5 cards can be drawn from a deck of 52 cards without replacement?
Tap to reveal 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.
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.
← Didn't Know|Knew It →
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
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
Tap to reveal 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.
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.
← Didn't Know|Knew It →
In how many ways can five different colored balls be arranged in a row?
In how many ways can five different colored balls be arranged in a row?
Tap to reveal 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.
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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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?
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?
Tap to reveal 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:
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
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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?
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?
Tap to reveal 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.
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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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.
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.
Tap to reveal 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.
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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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?
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?
Tap to reveal 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.
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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