Counting fruits
$begingroup$
Your local grocery store just received a large shipment of apples, oranges, pears, and bananas—more than 100 pieces each. You are shopping at the store and will purchase your fruit for the week.
- How many ways can you select 10 pieces of fruit from your store’s supply of apples, oranges, pears, and bananas?
Is it like there are more than 400 fruits in total. so C((400+10-1), (10-1))?
- How many ways can you select 10 pieces of fruit from your store’s supply of apples, oranges, pears, and bananas if you need at least one piece of each kind of fruit?
combinatorics
asked Dec 6 '18 at 16:20
CompComp
63
$endgroup$
add a comment |
$begingroup$
Your local grocery store just received a large shipment of apples, oranges, pears, and bananas—more than 100 pieces each. You are shopping at the store and will purchase your fruit for the week.
- How many ways can you select 10 pieces of fruit from your store’s supply of apples, oranges, pears, and bananas?
Is it like there are more than 400 fruits in total. so C((400+10-1), (10-1))?
- How many ways can you select 10 pieces of fruit from your store’s supply of apples, oranges, pears, and bananas if you need at least one piece of each kind of fruit?
combinatorics
asked Dec 6 '18 at 16:20
CompComp
63
$endgroup$
3
$begingroup$
No. This is a stars and bars problem.
$endgroup$
– saulspatz
Dec 6 '18 at 16:21
add a comment |
$begingroup$
Your local grocery store just received a large shipment of apples, oranges, pears, and bananas—more than 100 pieces each. You are shopping at the store and will purchase your fruit for the week.
- How many ways can you select 10 pieces of fruit from your store’s supply of apples, oranges, pears, and bananas?
Is it like there are more than 400 fruits in total. so C((400+10-1), (10-1))?
- How many ways can you select 10 pieces of fruit from your store’s supply of apples, oranges, pears, and bananas if you need at least one piece of each kind of fruit?
combinatorics
asked Dec 6 '18 at 16:20
CompComp
63
$endgroup$
Your local grocery store just received a large shipment of apples, oranges, pears, and bananas—more than 100 pieces each. You are shopping at the store and will purchase your fruit for the week.
- How many ways can you select 10 pieces of fruit from your store’s supply of apples, oranges, pears, and bananas?
Is it like there are more than 400 fruits in total. so C((400+10-1), (10-1))?
- How many ways can you select 10 pieces of fruit from your store’s supply of apples, oranges, pears, and bananas if you need at least one piece of each kind of fruit?
combinatorics
combinatorics
asked Dec 6 '18 at 16:20
CompComp
63
asked Dec 6 '18 at 16:20
CompComp
63
asked Dec 6 '18 at 16:20
CompComp
63
asked Dec 6 '18 at 16:20
CompComp
63
asked Dec 6 '18 at 16:20
CompComp
63
63
3
$begingroup$
No. This is a stars and bars problem.
$endgroup$
– saulspatz
Dec 6 '18 at 16:21
add a comment |
3
$begingroup$
No. This is a stars and bars problem.
$endgroup$
– saulspatz
Dec 6 '18 at 16:21
3
3
$begingroup$
No. This is a stars and bars problem.
$endgroup$
– saulspatz
Dec 6 '18 at 16:21
$begingroup$
No. This is a stars and bars problem.
$endgroup$
– saulspatz
Dec 6 '18 at 16:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As stated in the comment, this is a stars and bars problem. The supply of the grocery store is of no concern as long as they have more than your maximum amount for each fruit (10 of each fruit in this case)
Now let o denote a fruit and | denote a bracket. You'll have 10 o as you are going to pick 10 fruits. You'll need only 3 | tho, because 3 separation points are enough to divide into 4 groups.
Now that you have 10 fruits and 3 seperation points, each of the combinations will represent you how much to buy which fruit. for instance;
000|00|00|000 denotes you'll buy 3 of the first fruit 2 of the second 2 of the third and 3 of the 4th.
0000000||000| denotes that you'll get 7 of the first fruit , 3 of the thrid and none of the others.
As you can see we can arrange it in $frac {13!} {10!.3!}$ = $13 choose 3$ ways.
For the second part of the question, we need to pre-allocate 1 fruit each to all brackets, so you will have 4 of the 10 fruits already allocated, for the remaining 6 we can use the same method and find that we can choose in $6+3 choose 3$ = $9 choose 3 $ ways
answered Dec 6 '18 at 16:39
OfyaOfya
5048
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
As stated in the comment, this is a stars and bars problem. The supply of the grocery store is of no concern as long as they have more than your maximum amount for each fruit (10 of each fruit in this case)
Now let o denote a fruit and | denote a bracket. You'll have 10 o as you are going to pick 10 fruits. You'll need only 3 | tho, because 3 separation points are enough to divide into 4 groups.
Now that you have 10 fruits and 3 seperation points, each of the combinations will represent you how much to buy which fruit. for instance;
000|00|00|000 denotes you'll buy 3 of the first fruit 2 of the second 2 of the third and 3 of the 4th.
0000000||000| denotes that you'll get 7 of the first fruit , 3 of the thrid and none of the others.
As you can see we can arrange it in $frac {13!} {10!.3!}$ = $13 choose 3$ ways.
For the second part of the question, we need to pre-allocate 1 fruit each to all brackets, so you will have 4 of the 10 fruits already allocated, for the remaining 6 we can use the same method and find that we can choose in $6+3 choose 3$ = $9 choose 3 $ ways
answered Dec 6 '18 at 16:39
OfyaOfya
5048
$endgroup$
add a comment |
$begingroup$
As stated in the comment, this is a stars and bars problem. The supply of the grocery store is of no concern as long as they have more than your maximum amount for each fruit (10 of each fruit in this case)
Now let o denote a fruit and | denote a bracket. You'll have 10 o as you are going to pick 10 fruits. You'll need only 3 | tho, because 3 separation points are enough to divide into 4 groups.
Now that you have 10 fruits and 3 seperation points, each of the combinations will represent you how much to buy which fruit. for instance;
000|00|00|000 denotes you'll buy 3 of the first fruit 2 of the second 2 of the third and 3 of the 4th.
0000000||000| denotes that you'll get 7 of the first fruit , 3 of the thrid and none of the others.
As you can see we can arrange it in $frac {13!} {10!.3!}$ = $13 choose 3$ ways.
For the second part of the question, we need to pre-allocate 1 fruit each to all brackets, so you will have 4 of the 10 fruits already allocated, for the remaining 6 we can use the same method and find that we can choose in $6+3 choose 3$ = $9 choose 3 $ ways
answered Dec 6 '18 at 16:39
OfyaOfya
5048
$endgroup$
add a comment |
$begingroup$
As stated in the comment, this is a stars and bars problem. The supply of the grocery store is of no concern as long as they have more than your maximum amount for each fruit (10 of each fruit in this case)
Now let o denote a fruit and | denote a bracket. You'll have 10 o as you are going to pick 10 fruits. You'll need only 3 | tho, because 3 separation points are enough to divide into 4 groups.
Now that you have 10 fruits and 3 seperation points, each of the combinations will represent you how much to buy which fruit. for instance;
000|00|00|000 denotes you'll buy 3 of the first fruit 2 of the second 2 of the third and 3 of the 4th.
0000000||000| denotes that you'll get 7 of the first fruit , 3 of the thrid and none of the others.
As you can see we can arrange it in $frac {13!} {10!.3!}$ = $13 choose 3$ ways.
For the second part of the question, we need to pre-allocate 1 fruit each to all brackets, so you will have 4 of the 10 fruits already allocated, for the remaining 6 we can use the same method and find that we can choose in $6+3 choose 3$ = $9 choose 3 $ ways
answered Dec 6 '18 at 16:39
OfyaOfya
5048
$endgroup$
As stated in the comment, this is a stars and bars problem. The supply of the grocery store is of no concern as long as they have more than your maximum amount for each fruit (10 of each fruit in this case)
Now let o denote a fruit and | denote a bracket. You'll have 10 o as you are going to pick 10 fruits. You'll need only 3 | tho, because 3 separation points are enough to divide into 4 groups.
Now that you have 10 fruits and 3 seperation points, each of the combinations will represent you how much to buy which fruit. for instance;
000|00|00|000 denotes you'll buy 3 of the first fruit 2 of the second 2 of the third and 3 of the 4th.
0000000||000| denotes that you'll get 7 of the first fruit , 3 of the thrid and none of the others.
As you can see we can arrange it in $frac {13!} {10!.3!}$ = $13 choose 3$ ways.
For the second part of the question, we need to pre-allocate 1 fruit each to all brackets, so you will have 4 of the 10 fruits already allocated, for the remaining 6 we can use the same method and find that we can choose in $6+3 choose 3$ = $9 choose 3 $ ways
answered Dec 6 '18 at 16:39
OfyaOfya
5048
answered Dec 6 '18 at 16:39
OfyaOfya
5048
answered Dec 6 '18 at 16:39
OfyaOfya
5048
answered Dec 6 '18 at 16:39
OfyaOfya
5048
5048
add a comment |
add a comment |
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3
$begingroup$
No. This is a stars and bars problem.
$endgroup$
– saulspatz
Dec 6 '18 at 16:21