Filling a barrel, using small containers












0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 22:34









Gerry Myerson

146k8147298




146k8147298










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














  • 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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $b$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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1 Answer
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active

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active

oldest

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2












$begingroup$

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 $b$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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 $b$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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 $b$.






share|cite|improve this answer









$endgroup$



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 $b$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 9:39









gandalf61gandalf61

8,076625




8,076625












  • $begingroup$
    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










  • $begingroup$
    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


















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