Discrete math counting principles
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In how many ways can a teacher distribute 12 identical science books among 15 students if
1) no student gets more than one book?
2) if the oldest student gets two books but no other student gets more than one book?
Initially, I tried the binomial theorems with the idea that the student gets a book or no book but that is not giving me the correct solution.
combinatorics permutations combinations
edited Dec 9 '18 at 19:44
Robert Z
97.4k1066137
asked Dec 9 '18 at 19:38
critter1854critter1854
1
$endgroup$
add a comment |
$begingroup$
In how many ways can a teacher distribute 12 identical science books among 15 students if
1) no student gets more than one book?
2) if the oldest student gets two books but no other student gets more than one book?
Initially, I tried the binomial theorems with the idea that the student gets a book or no book but that is not giving me the correct solution.
combinatorics permutations combinations
edited Dec 9 '18 at 19:44
Robert Z
97.4k1066137
asked Dec 9 '18 at 19:38
critter1854critter1854
1
$endgroup$
add a comment |
$begingroup$
In how many ways can a teacher distribute 12 identical science books among 15 students if
1) no student gets more than one book?
2) if the oldest student gets two books but no other student gets more than one book?
Initially, I tried the binomial theorems with the idea that the student gets a book or no book but that is not giving me the correct solution.
combinatorics permutations combinations
edited Dec 9 '18 at 19:44
Robert Z
97.4k1066137
asked Dec 9 '18 at 19:38
critter1854critter1854
1
$endgroup$
In how many ways can a teacher distribute 12 identical science books among 15 students if
1) no student gets more than one book?
2) if the oldest student gets two books but no other student gets more than one book?
Initially, I tried the binomial theorems with the idea that the student gets a book or no book but that is not giving me the correct solution.
combinatorics permutations combinations
combinatorics permutations combinations
edited Dec 9 '18 at 19:44
Robert Z
97.4k1066137
asked Dec 9 '18 at 19:38
critter1854critter1854
1
edited Dec 9 '18 at 19:44
Robert Z
97.4k1066137
asked Dec 9 '18 at 19:38
critter1854critter1854
1
edited Dec 9 '18 at 19:44
Robert Z
97.4k1066137
edited Dec 9 '18 at 19:44
Robert Z
97.4k1066137
edited Dec 9 '18 at 19:44
Robert Z
97.4k1066137
97.4k1066137
asked Dec 9 '18 at 19:38
critter1854critter1854
1
asked Dec 9 '18 at 19:38
critter1854critter1854
1
asked Dec 9 '18 at 19:38
critter1854critter1854
1
1
add a comment |
add a comment |
2 Answers
2
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oldest
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Hint for 1) Count the number of subsets of cardinality $12$ of a set of $15$ elements.
Hint for 2) Count the number of subsets of cardinality $12-2$ of a set of $15-1$ elements.
answered Dec 9 '18 at 19:47
Robert ZRobert Z
97.4k1066137
$endgroup$
$begingroup$
The books are identical!
$endgroup$
– Robert Z
Dec 9 '18 at 20:39
add a comment |
$begingroup$
Assuming the students are distinguishable:
For the first one: If no student gets more than one book, then it is just a matter of finding the number of ways to pick 12 students out of those 15, i.e. $15 choose 12$
For the second one: the oldest one gets 2 books ... leaving 14 students ... and 10 books ... can you do the rest?
answered Dec 9 '18 at 19:45
Bram28Bram28
62.4k44793
$endgroup$
$begingroup$
This is an exam prep so I don't have the actual answers, so I just want to make sure I am correct in my thinking. So the first is 15c12 = 455 and the second is 14c10 =1001
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– critter1854
Dec 9 '18 at 19:58
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@critter1854 Yes, correct!
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– Bram28
Dec 9 '18 at 20:07
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@N.F.Taussig But the books are identical ...
$endgroup$
– Bram28
Dec 9 '18 at 20:31
add a comment |
Your Answer
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2 Answers
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oldest
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2 Answers
2
active
oldest
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$begingroup$
Hint for 1) Count the number of subsets of cardinality $12$ of a set of $15$ elements.
Hint for 2) Count the number of subsets of cardinality $12-2$ of a set of $15-1$ elements.
answered Dec 9 '18 at 19:47
Robert ZRobert Z
97.4k1066137
$endgroup$
$begingroup$
The books are identical!
$endgroup$
– Robert Z
Dec 9 '18 at 20:39
add a comment |
$begingroup$
Hint for 1) Count the number of subsets of cardinality $12$ of a set of $15$ elements.
Hint for 2) Count the number of subsets of cardinality $12-2$ of a set of $15-1$ elements.
answered Dec 9 '18 at 19:47
Robert ZRobert Z
97.4k1066137
$endgroup$
$begingroup$
The books are identical!
$endgroup$
– Robert Z
Dec 9 '18 at 20:39
add a comment |
$begingroup$
Hint for 1) Count the number of subsets of cardinality $12$ of a set of $15$ elements.
Hint for 2) Count the number of subsets of cardinality $12-2$ of a set of $15-1$ elements.
answered Dec 9 '18 at 19:47
Robert ZRobert Z
97.4k1066137
$endgroup$
Hint for 1) Count the number of subsets of cardinality $12$ of a set of $15$ elements.
Hint for 2) Count the number of subsets of cardinality $12-2$ of a set of $15-1$ elements.
answered Dec 9 '18 at 19:47
Robert ZRobert Z
97.4k1066137
answered Dec 9 '18 at 19:47
Robert ZRobert Z
97.4k1066137
answered Dec 9 '18 at 19:47
Robert ZRobert Z
97.4k1066137
answered Dec 9 '18 at 19:47
Robert ZRobert Z
97.4k1066137
97.4k1066137
$begingroup$
The books are identical!
$endgroup$
– Robert Z
Dec 9 '18 at 20:39
add a comment |
$begingroup$
The books are identical!
$endgroup$
– Robert Z
Dec 9 '18 at 20:39
$begingroup$
The books are identical!
$endgroup$
– Robert Z
Dec 9 '18 at 20:39
$begingroup$
The books are identical!
$endgroup$
– Robert Z
Dec 9 '18 at 20:39
add a comment |
$begingroup$
Assuming the students are distinguishable:
For the first one: If no student gets more than one book, then it is just a matter of finding the number of ways to pick 12 students out of those 15, i.e. $15 choose 12$
For the second one: the oldest one gets 2 books ... leaving 14 students ... and 10 books ... can you do the rest?
answered Dec 9 '18 at 19:45
Bram28Bram28
62.4k44793
$endgroup$
$begingroup$
This is an exam prep so I don't have the actual answers, so I just want to make sure I am correct in my thinking. So the first is 15c12 = 455 and the second is 14c10 =1001
$endgroup$
– critter1854
Dec 9 '18 at 19:58
$begingroup$
@critter1854 Yes, correct!
$endgroup$
– Bram28
Dec 9 '18 at 20:07
$begingroup$
@N.F.Taussig But the books are identical ...
$endgroup$
– Bram28
Dec 9 '18 at 20:31
add a comment |
$begingroup$
Assuming the students are distinguishable:
For the first one: If no student gets more than one book, then it is just a matter of finding the number of ways to pick 12 students out of those 15, i.e. $15 choose 12$
For the second one: the oldest one gets 2 books ... leaving 14 students ... and 10 books ... can you do the rest?
answered Dec 9 '18 at 19:45
Bram28Bram28
62.4k44793
$endgroup$
$begingroup$
This is an exam prep so I don't have the actual answers, so I just want to make sure I am correct in my thinking. So the first is 15c12 = 455 and the second is 14c10 =1001
$endgroup$
– critter1854
Dec 9 '18 at 19:58
$begingroup$
@critter1854 Yes, correct!
$endgroup$
– Bram28
Dec 9 '18 at 20:07
$begingroup$
@N.F.Taussig But the books are identical ...
$endgroup$
– Bram28
Dec 9 '18 at 20:31
add a comment |
$begingroup$
Assuming the students are distinguishable:
For the first one: If no student gets more than one book, then it is just a matter of finding the number of ways to pick 12 students out of those 15, i.e. $15 choose 12$
For the second one: the oldest one gets 2 books ... leaving 14 students ... and 10 books ... can you do the rest?
answered Dec 9 '18 at 19:45
Bram28Bram28
62.4k44793
$endgroup$
Assuming the students are distinguishable:
For the first one: If no student gets more than one book, then it is just a matter of finding the number of ways to pick 12 students out of those 15, i.e. $15 choose 12$
For the second one: the oldest one gets 2 books ... leaving 14 students ... and 10 books ... can you do the rest?
answered Dec 9 '18 at 19:45
Bram28Bram28
62.4k44793
edited Dec 9 '18 at 19:50
answered Dec 9 '18 at 19:45
Bram28Bram28
62.4k44793
answered Dec 9 '18 at 19:45
Bram28Bram28
62.4k44793
answered Dec 9 '18 at 19:45
Bram28Bram28
62.4k44793
62.4k44793
$begingroup$
This is an exam prep so I don't have the actual answers, so I just want to make sure I am correct in my thinking. So the first is 15c12 = 455 and the second is 14c10 =1001
$endgroup$
– critter1854
Dec 9 '18 at 19:58
$begingroup$
@critter1854 Yes, correct!
$endgroup$
– Bram28
Dec 9 '18 at 20:07
$begingroup$
@N.F.Taussig But the books are identical ...
$endgroup$
– Bram28
Dec 9 '18 at 20:31
add a comment |
$begingroup$
This is an exam prep so I don't have the actual answers, so I just want to make sure I am correct in my thinking. So the first is 15c12 = 455 and the second is 14c10 =1001
$endgroup$
– critter1854
Dec 9 '18 at 19:58
$begingroup$
@critter1854 Yes, correct!
$endgroup$
– Bram28
Dec 9 '18 at 20:07
$begingroup$
@N.F.Taussig But the books are identical ...
$endgroup$
– Bram28
Dec 9 '18 at 20:31
$begingroup$
This is an exam prep so I don't have the actual answers, so I just want to make sure I am correct in my thinking. So the first is 15c12 = 455 and the second is 14c10 =1001
$endgroup$
– critter1854
Dec 9 '18 at 19:58
$begingroup$
This is an exam prep so I don't have the actual answers, so I just want to make sure I am correct in my thinking. So the first is 15c12 = 455 and the second is 14c10 =1001
$endgroup$
– critter1854
Dec 9 '18 at 19:58
$begingroup$
@critter1854 Yes, correct!
$endgroup$
– Bram28
Dec 9 '18 at 20:07
$begingroup$
@critter1854 Yes, correct!
$endgroup$
– Bram28
Dec 9 '18 at 20:07
$begingroup$
@N.F.Taussig But the books are identical ...
$endgroup$
– Bram28
Dec 9 '18 at 20:31
$begingroup$
@N.F.Taussig But the books are identical ...
$endgroup$
– Bram28
Dec 9 '18 at 20:31
add a comment |
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