Filling a barrel, using small containers
$begingroup$
In how many ways, using containers, one with 2 liters and other with 7 liters, can you fill a barrel of 1234 liters?
What's the fastest and what's the slowest way to fill the barrel?
Should I use formula for permutation n!/(n−r)! or the other formula n!/r!-(n−r)! . In similar tasks I don't know which formula should be used.
Can anyone help? Thank you in advance!
permutations combinations
edited Dec 4 '18 at 22:34
Gerry Myerson
146k8147298
asked Dec 4 '18 at 9:11
HausHaus
307
$endgroup$
|
show 1 more comment
$begingroup$
In how many ways, using containers, one with 2 liters and other with 7 liters, can you fill a barrel of 1234 liters?
What's the fastest and what's the slowest way to fill the barrel?
Should I use formula for permutation n!/(n−r)! or the other formula n!/r!-(n−r)! . In similar tasks I don't know which formula should be used.
Can anyone help? Thank you in advance!
permutations combinations
edited Dec 4 '18 at 22:34
Gerry Myerson
146k8147298
asked Dec 4 '18 at 9:11
HausHaus
307
$endgroup$
1
$begingroup$
So you're filling some 2 litre barrels and then some 7 little barrels, and pouring them together to make 1234 litres. Does the order matter? If not, then you're essentially looking for the number of integer solutions to $$ 2a + 7b = 1234 $$
$endgroup$
– Matti P.
Dec 4 '18 at 9:16
$begingroup$
Order does not metter. Using that equation how can I find what's the fastest and what's the slowest way to fill the barrel?
$endgroup$
– Haus
Dec 4 '18 at 9:19
$begingroup$
Since your problem definition is not mentioning anything about time (for ex, how long it takes to fill a 2 / 7 litre barrel, or how long it takes to pour a barrel into the big barrel), it's a bit tricky to model the problem that way.
$endgroup$
– Matti P.
Dec 4 '18 at 9:20
1
$begingroup$
Some thoughts: Since $2a+7b = 1234$, we can take mod 7 on both sides, resulting in $$ 2a equiv 2 Rightarrow a equiv 1 ~text{mod} 7 $$ In addition, $a geq 0$ and $a leq frac{1234}{2} = 617$. Then, how many numbers that are $equiv 1~text{mod}~7$ between $0$ and $617$ ?
$endgroup$
– Matti P.
Dec 4 '18 at 9:26
$begingroup$
I thought solving this as Diophantine equation.
$endgroup$
– Haus
Dec 4 '18 at 9:30
|
show 1 more comment
$begingroup$
In how many ways, using containers, one with 2 liters and other with 7 liters, can you fill a barrel of 1234 liters?
What's the fastest and what's the slowest way to fill the barrel?
Should I use formula for permutation n!/(n−r)! or the other formula n!/r!-(n−r)! . In similar tasks I don't know which formula should be used.
Can anyone help? Thank you in advance!
permutations combinations
edited Dec 4 '18 at 22:34
Gerry Myerson
146k8147298
asked Dec 4 '18 at 9:11
HausHaus
307
$endgroup$
In how many ways, using containers, one with 2 liters and other with 7 liters, can you fill a barrel of 1234 liters?
What's the fastest and what's the slowest way to fill the barrel?
Should I use formula for permutation n!/(n−r)! or the other formula n!/r!-(n−r)! . In similar tasks I don't know which formula should be used.
Can anyone help? Thank you in advance!
permutations combinations
permutations combinations
edited Dec 4 '18 at 22:34
Gerry Myerson
146k8147298
asked Dec 4 '18 at 9:11
HausHaus
307
edited Dec 4 '18 at 22:34
Gerry Myerson
146k8147298
asked Dec 4 '18 at 9:11
HausHaus
307
edited Dec 4 '18 at 22:34
Gerry Myerson
146k8147298
edited Dec 4 '18 at 22:34
Gerry Myerson
146k8147298
edited Dec 4 '18 at 22:34
Gerry Myerson
146k8147298
146k8147298
asked Dec 4 '18 at 9:11
HausHaus
307
asked Dec 4 '18 at 9:11
HausHaus
307
asked Dec 4 '18 at 9:11
HausHaus
307
307
1
$begingroup$
So you're filling some 2 litre barrels and then some 7 little barrels, and pouring them together to make 1234 litres. Does the order matter? If not, then you're essentially looking for the number of integer solutions to $$ 2a + 7b = 1234 $$
$endgroup$
– Matti P.
Dec 4 '18 at 9:16
$begingroup$
Order does not metter. Using that equation how can I find what's the fastest and what's the slowest way to fill the barrel?
$endgroup$
– Haus
Dec 4 '18 at 9:19
$begingroup$
Since your problem definition is not mentioning anything about time (for ex, how long it takes to fill a 2 / 7 litre barrel, or how long it takes to pour a barrel into the big barrel), it's a bit tricky to model the problem that way.
$endgroup$
– Matti P.
Dec 4 '18 at 9:20
1
$begingroup$
Some thoughts: Since $2a+7b = 1234$, we can take mod 7 on both sides, resulting in $$ 2a equiv 2 Rightarrow a equiv 1 ~text{mod} 7 $$ In addition, $a geq 0$ and $a leq frac{1234}{2} = 617$. Then, how many numbers that are $equiv 1~text{mod}~7$ between $0$ and $617$ ?
$endgroup$
– Matti P.
Dec 4 '18 at 9:26
$begingroup$
I thought solving this as Diophantine equation.
$endgroup$
– Haus
Dec 4 '18 at 9:30
|
show 1 more comment
1
$begingroup$
So you're filling some 2 litre barrels and then some 7 little barrels, and pouring them together to make 1234 litres. Does the order matter? If not, then you're essentially looking for the number of integer solutions to $$ 2a + 7b = 1234 $$
$endgroup$
– Matti P.
Dec 4 '18 at 9:16
$begingroup$
Order does not metter. Using that equation how can I find what's the fastest and what's the slowest way to fill the barrel?
$endgroup$
– Haus
Dec 4 '18 at 9:19
$begingroup$
Since your problem definition is not mentioning anything about time (for ex, how long it takes to fill a 2 / 7 litre barrel, or how long it takes to pour a barrel into the big barrel), it's a bit tricky to model the problem that way.
$endgroup$
– Matti P.
Dec 4 '18 at 9:20
1
$begingroup$
Some thoughts: Since $2a+7b = 1234$, we can take mod 7 on both sides, resulting in $$ 2a equiv 2 Rightarrow a equiv 1 ~text{mod} 7 $$ In addition, $a geq 0$ and $a leq frac{1234}{2} = 617$. Then, how many numbers that are $equiv 1~text{mod}~7$ between $0$ and $617$ ?
$endgroup$
– Matti P.
Dec 4 '18 at 9:26
$begingroup$
I thought solving this as Diophantine equation.
$endgroup$
– Haus
Dec 4 '18 at 9:30
1
1
$begingroup$
So you're filling some 2 litre barrels and then some 7 little barrels, and pouring them together to make 1234 litres. Does the order matter? If not, then you're essentially looking for the number of integer solutions to $$ 2a + 7b = 1234 $$
$endgroup$
– Matti P.
Dec 4 '18 at 9:16
$begingroup$
So you're filling some 2 litre barrels and then some 7 little barrels, and pouring them together to make 1234 litres. Does the order matter? If not, then you're essentially looking for the number of integer solutions to $$ 2a + 7b = 1234 $$
$endgroup$
– Matti P.
Dec 4 '18 at 9:16
$begingroup$
Order does not metter. Using that equation how can I find what's the fastest and what's the slowest way to fill the barrel?
$endgroup$
– Haus
Dec 4 '18 at 9:19
$begingroup$
Order does not metter. Using that equation how can I find what's the fastest and what's the slowest way to fill the barrel?
$endgroup$
– Haus
Dec 4 '18 at 9:19
$begingroup$
Since your problem definition is not mentioning anything about time (for ex, how long it takes to fill a 2 / 7 litre barrel, or how long it takes to pour a barrel into the big barrel), it's a bit tricky to model the problem that way.
$endgroup$
– Matti P.
Dec 4 '18 at 9:20
$begingroup$
Since your problem definition is not mentioning anything about time (for ex, how long it takes to fill a 2 / 7 litre barrel, or how long it takes to pour a barrel into the big barrel), it's a bit tricky to model the problem that way.
$endgroup$
– Matti P.
Dec 4 '18 at 9:20
1
1
$begingroup$
Some thoughts: Since $2a+7b = 1234$, we can take mod 7 on both sides, resulting in $$ 2a equiv 2 Rightarrow a equiv 1 ~text{mod} 7 $$ In addition, $a geq 0$ and $a leq frac{1234}{2} = 617$. Then, how many numbers that are $equiv 1~text{mod}~7$ between $0$ and $617$ ?
$endgroup$
– Matti P.
Dec 4 '18 at 9:26
$begingroup$
Some thoughts: Since $2a+7b = 1234$, we can take mod 7 on both sides, resulting in $$ 2a equiv 2 Rightarrow a equiv 1 ~text{mod} 7 $$ In addition, $a geq 0$ and $a leq frac{1234}{2} = 617$. Then, how many numbers that are $equiv 1~text{mod}~7$ between $0$ and $617$ ?
$endgroup$
– Matti P.
Dec 4 '18 at 9:26
$begingroup$
I thought solving this as Diophantine equation.
$endgroup$
– Haus
Dec 4 '18 at 9:30
$begingroup$
I thought solving this as Diophantine equation.
$endgroup$
– Haus
Dec 4 '18 at 9:30
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If the order of using the $2$ and $7$ liter barrels does not matter then you do not need combinations or permutations. You simply need to find the number of solutions to
$2a + 7b = 1234$
where $a$ and $b$ are non-negative integers (you can use $0$ barrels of either size, but you cannot use $-1$ barrels).
So the minimum value of $a$ is $0$. And the maximum value is $frac{1234}{2}=617$. So we have
$0 le a le 617$
In addition, to make $7b=1234 - 2a$ a multiple of $7$, we must have $a mod 7 = 1$.
So $a$ can be $1, 8, 15, 22, dots, 610, 617$. The corresponding values of $b$ are $176, 174, 172, 170, dots, 2, 0$.
From this you should be able to work out how many possible values there are for $a$ and $b$.
answered Dec 4 '18 at 9:39
gandalf61gandalf61
8,011625
$endgroup$
$begingroup$
How many possible values for a and b, do i use the formula for binomial coefficient? Thank you.
$endgroup$
– Haus
Dec 4 '18 at 9:44
$begingroup$
You are over-complicating this. Look at the possible values for $b$. How many integers between $0$ and $176$ inclusive are multiples of $2$ ?
$endgroup$
– gandalf61
Dec 4 '18 at 10:21
add a comment |
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$begingroup$
If the order of using the $2$ and $7$ liter barrels does not matter then you do not need combinations or permutations. You simply need to find the number of solutions to
$2a + 7b = 1234$
where $a$ and $b$ are non-negative integers (you can use $0$ barrels of either size, but you cannot use $-1$ barrels).
So the minimum value of $a$ is $0$. And the maximum value is $frac{1234}{2}=617$. So we have
$0 le a le 617$
In addition, to make $7b=1234 - 2a$ a multiple of $7$, we must have $a mod 7 = 1$.
So $a$ can be $1, 8, 15, 22, dots, 610, 617$. The corresponding values of $b$ are $176, 174, 172, 170, dots, 2, 0$.
From this you should be able to work out how many possible values there are for $a$ and $b$.
answered Dec 4 '18 at 9:39
gandalf61gandalf61
8,011625
$endgroup$
$begingroup$
How many possible values for a and b, do i use the formula for binomial coefficient? Thank you.
$endgroup$
– Haus
Dec 4 '18 at 9:44
$begingroup$
You are over-complicating this. Look at the possible values for $b$. How many integers between $0$ and $176$ inclusive are multiples of $2$ ?
$endgroup$
– gandalf61
Dec 4 '18 at 10:21
add a comment |
$begingroup$
If the order of using the $2$ and $7$ liter barrels does not matter then you do not need combinations or permutations. You simply need to find the number of solutions to
$2a + 7b = 1234$
where $a$ and $b$ are non-negative integers (you can use $0$ barrels of either size, but you cannot use $-1$ barrels).
So the minimum value of $a$ is $0$. And the maximum value is $frac{1234}{2}=617$. So we have
$0 le a le 617$
In addition, to make $7b=1234 - 2a$ a multiple of $7$, we must have $a mod 7 = 1$.
So $a$ can be $1, 8, 15, 22, dots, 610, 617$. The corresponding values of $b$ are $176, 174, 172, 170, dots, 2, 0$.
From this you should be able to work out how many possible values there are for $a$ and $b$.
answered Dec 4 '18 at 9:39
gandalf61gandalf61
8,011625
$endgroup$
$begingroup$
How many possible values for a and b, do i use the formula for binomial coefficient? Thank you.
$endgroup$
– Haus
Dec 4 '18 at 9:44
$begingroup$
You are over-complicating this. Look at the possible values for $b$. How many integers between $0$ and $176$ inclusive are multiples of $2$ ?
$endgroup$
– gandalf61
Dec 4 '18 at 10:21
add a comment |
$begingroup$
If the order of using the $2$ and $7$ liter barrels does not matter then you do not need combinations or permutations. You simply need to find the number of solutions to
$2a + 7b = 1234$
where $a$ and $b$ are non-negative integers (you can use $0$ barrels of either size, but you cannot use $-1$ barrels).
So the minimum value of $a$ is $0$. And the maximum value is $frac{1234}{2}=617$. So we have
$0 le a le 617$
In addition, to make $7b=1234 - 2a$ a multiple of $7$, we must have $a mod 7 = 1$.
So $a$ can be $1, 8, 15, 22, dots, 610, 617$. The corresponding values of $b$ are $176, 174, 172, 170, dots, 2, 0$.
From this you should be able to work out how many possible values there are for $a$ and $b$.
answered Dec 4 '18 at 9:39
gandalf61gandalf61
8,011625
$endgroup$
If the order of using the $2$ and $7$ liter barrels does not matter then you do not need combinations or permutations. You simply need to find the number of solutions to
$2a + 7b = 1234$
where $a$ and $b$ are non-negative integers (you can use $0$ barrels of either size, but you cannot use $-1$ barrels).
So the minimum value of $a$ is $0$. And the maximum value is $frac{1234}{2}=617$. So we have
$0 le a le 617$
In addition, to make $7b=1234 - 2a$ a multiple of $7$, we must have $a mod 7 = 1$.
So $a$ can be $1, 8, 15, 22, dots, 610, 617$. The corresponding values of $b$ are $176, 174, 172, 170, dots, 2, 0$.
From this you should be able to work out how many possible values there are for $a$ and $b$.
answered Dec 4 '18 at 9:39
gandalf61gandalf61
8,011625
answered Dec 4 '18 at 9:39
gandalf61gandalf61
8,011625
answered Dec 4 '18 at 9:39
gandalf61gandalf61
8,011625
answered Dec 4 '18 at 9:39
gandalf61gandalf61
8,011625
8,011625
$begingroup$
How many possible values for a and b, do i use the formula for binomial coefficient? Thank you.
$endgroup$
– Haus
Dec 4 '18 at 9:44
$begingroup$
You are over-complicating this. Look at the possible values for $b$. How many integers between $0$ and $176$ inclusive are multiples of $2$ ?
$endgroup$
– gandalf61
Dec 4 '18 at 10:21
add a comment |
$begingroup$
How many possible values for a and b, do i use the formula for binomial coefficient? Thank you.
$endgroup$
– Haus
Dec 4 '18 at 9:44
$begingroup$
You are over-complicating this. Look at the possible values for $b$. How many integers between $0$ and $176$ inclusive are multiples of $2$ ?
$endgroup$
– gandalf61
Dec 4 '18 at 10:21
$begingroup$
How many possible values for a and b, do i use the formula for binomial coefficient? Thank you.
$endgroup$
– Haus
Dec 4 '18 at 9:44
$begingroup$
How many possible values for a and b, do i use the formula for binomial coefficient? Thank you.
$endgroup$
– Haus
Dec 4 '18 at 9:44
$begingroup$
You are over-complicating this. Look at the possible values for $b$. How many integers between $0$ and $176$ inclusive are multiples of $2$ ?
$endgroup$
– gandalf61
Dec 4 '18 at 10:21
$begingroup$
You are over-complicating this. Look at the possible values for $b$. How many integers between $0$ and $176$ inclusive are multiples of $2$ ?
$endgroup$
– gandalf61
Dec 4 '18 at 10:21
add a comment |
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1
$begingroup$
So you're filling some 2 litre barrels and then some 7 little barrels, and pouring them together to make 1234 litres. Does the order matter? If not, then you're essentially looking for the number of integer solutions to $$ 2a + 7b = 1234 $$
$endgroup$
– Matti P.
Dec 4 '18 at 9:16
$begingroup$
Order does not metter. Using that equation how can I find what's the fastest and what's the slowest way to fill the barrel?
$endgroup$
– Haus
Dec 4 '18 at 9:19
$begingroup$
Since your problem definition is not mentioning anything about time (for ex, how long it takes to fill a 2 / 7 litre barrel, or how long it takes to pour a barrel into the big barrel), it's a bit tricky to model the problem that way.
$endgroup$
– Matti P.
Dec 4 '18 at 9:20
1
$begingroup$
Some thoughts: Since $2a+7b = 1234$, we can take mod 7 on both sides, resulting in $$ 2a equiv 2 Rightarrow a equiv 1 ~text{mod} 7 $$ In addition, $a geq 0$ and $a leq frac{1234}{2} = 617$. Then, how many numbers that are $equiv 1~text{mod}~7$ between $0$ and $617$ ?
$endgroup$
– Matti P.
Dec 4 '18 at 9:26
$begingroup$
I thought solving this as Diophantine equation.
$endgroup$
– Haus
Dec 4 '18 at 9:30