user11235813
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- Sep 28, 2021
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An encyclopedia consists of 8 volumes, numbered 1 to 8, initially arranged in ascending order of their numbers.
a) How many ways can we put these volumes on a shelf?
b) In how many ways can we put these volumes on a shelf so that at least one volume is not occupying the same starting position?
c) In how many ways can we place these volumes on a shelf so that exactly one volume is not placed in ascending order? For example, 2, 1, 3, 4, 5, 6, 7, 8 or 1, 2, 4, 5, 3, 6, 7, 8.
d) In how many ways can we select 3 of the 8 volumes so that they do not have consecutive numbers?
Book answers:
a) 8! b) 8!−1 c) 8⋅7−7 or 7⋅7 d) C(6,3)
My answers:
a) Permutation of 8 different volumes in a row: 8! (correct)
b) I fixed the 1 in first place and changed the other 7, because within the possibilities, only the 1 can be in its initial place with all out of their respective initial places, or all can match in their respective initial places as well, as the utterance asks for at least 1 in the starting place then it would, by indirect count: 8!−7! (but incorrect, I don't understand why it subtracts only one possibility if one of them would be in the case that they all coincided in the starting position, but it doesn't necessarily have to happen, as the statement says at least one)
c) As it's just one number out of place, keeping everyone in their starting position and choosing 1 of the 8 to put between them we have 7 possibilities to allocate with 8 possible numbers: 8⋅7 (but incorrect, I don't understand why it subtracts 7 from this count)
d) As it must not be the 3 consecutive ones, for one of the numbers in the count we cannot count its smallest consecutive and its biggest consecutive must not be included in the count, so as there are 8 numbers we will only have 6 to choose 3: C(6,3), but as we need to choose 6 out of 8 totals possibles so getting C(6,3)⋅C(8,6) (but also incorrect)
What did I think wrong and how should I have thought?
a) How many ways can we put these volumes on a shelf?
b) In how many ways can we put these volumes on a shelf so that at least one volume is not occupying the same starting position?
c) In how many ways can we place these volumes on a shelf so that exactly one volume is not placed in ascending order? For example, 2, 1, 3, 4, 5, 6, 7, 8 or 1, 2, 4, 5, 3, 6, 7, 8.
d) In how many ways can we select 3 of the 8 volumes so that they do not have consecutive numbers?
Book answers:
a) 8! b) 8!−1 c) 8⋅7−7 or 7⋅7 d) C(6,3)
My answers:
a) Permutation of 8 different volumes in a row: 8! (correct)
b) I fixed the 1 in first place and changed the other 7, because within the possibilities, only the 1 can be in its initial place with all out of their respective initial places, or all can match in their respective initial places as well, as the utterance asks for at least 1 in the starting place then it would, by indirect count: 8!−7! (but incorrect, I don't understand why it subtracts only one possibility if one of them would be in the case that they all coincided in the starting position, but it doesn't necessarily have to happen, as the statement says at least one)
c) As it's just one number out of place, keeping everyone in their starting position and choosing 1 of the 8 to put between them we have 7 possibilities to allocate with 8 possible numbers: 8⋅7 (but incorrect, I don't understand why it subtracts 7 from this count)
d) As it must not be the 3 consecutive ones, for one of the numbers in the count we cannot count its smallest consecutive and its biggest consecutive must not be included in the count, so as there are 8 numbers we will only have 6 to choose 3: C(6,3), but as we need to choose 6 out of 8 totals possibles so getting C(6,3)⋅C(8,6) (but also incorrect)
What did I think wrong and how should I have thought?
