Mixing
How many ways to pick 11 coins from those five piles of coins which must consists of all five different value...
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If we have a piles of 10 pennies , a pile of 10 nickels, a pile of 10 dimes , a pile of 9 quarters and a pile of 8 half-dollar coins.
How many ways to pick 11 coins from those five piles of coins which must consists of all five different value of coins?
I figured out how to pick 11 coins from five piles of coins:
(5-1 +11)
C
(11) = 15 C 11 = 1365
combinatorics discrete-mathematics combinations
asked yesterday
Mark
11
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up vote
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If we have a piles of 10 pennies , a pile of 10 nickels, a pile of 10 dimes , a pile of 9 quarters and a pile of 8 half-dollar coins.
How many ways to pick 11 coins from those five piles of coins which must consists of all five different value of coins?
I figured out how to pick 11 coins from five piles of coins:
(5-1 +11)
C
(11) = 15 C 11 = 1365
combinatorics discrete-mathematics combinations
asked yesterday
Mark
11
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. Usually people don’t answer to questions that show no effort on the questioner’s part because it makes the questioner appear to use this site as a homework help site, which it is not. Please edit your question to show more effort, such as any attempts.
– Kyky
yesterday
Chances are that you've already seen problems equivalent to this one and have been given a method to solve them. But the wording of those other problems may have been quite different. So it's a matter of trying to find the equivalence to a problem whose solution you already know.
– David K
yesterday
I have done the first part and edit it.
– Mark
yesterday
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
yesterday
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
22 hours ago
add a comment |
up vote
0
down vote
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up vote
0
down vote
favorite
If we have a piles of 10 pennies , a pile of 10 nickels, a pile of 10 dimes , a pile of 9 quarters and a pile of 8 half-dollar coins.
How many ways to pick 11 coins from those five piles of coins which must consists of all five different value of coins?
I figured out how to pick 11 coins from five piles of coins:
(5-1 +11)
C
(11) = 15 C 11 = 1365
combinatorics discrete-mathematics combinations
asked yesterday
Mark
11
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If we have a piles of 10 pennies , a pile of 10 nickels, a pile of 10 dimes , a pile of 9 quarters and a pile of 8 half-dollar coins.
How many ways to pick 11 coins from those five piles of coins which must consists of all five different value of coins?
I figured out how to pick 11 coins from five piles of coins:
(5-1 +11)
C
(11) = 15 C 11 = 1365
combinatorics discrete-mathematics combinations
combinatorics discrete-mathematics combinations
asked yesterday
Mark
11
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Mark
11
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 22 hours ago
asked yesterday
Mark
11
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Mark
11
asked yesterday
Mark
11
11
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. Usually people don’t answer to questions that show no effort on the questioner’s part because it makes the questioner appear to use this site as a homework help site, which it is not. Please edit your question to show more effort, such as any attempts.
– Kyky
yesterday
Chances are that you've already seen problems equivalent to this one and have been given a method to solve them. But the wording of those other problems may have been quite different. So it's a matter of trying to find the equivalence to a problem whose solution you already know.
– David K
yesterday
I have done the first part and edit it.
– Mark
yesterday
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
yesterday
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
22 hours ago
add a comment |
Welcome to MSE. Usually people don’t answer to questions that show no effort on the questioner’s part because it makes the questioner appear to use this site as a homework help site, which it is not. Please edit your question to show more effort, such as any attempts.
– Kyky
yesterday
Chances are that you've already seen problems equivalent to this one and have been given a method to solve them. But the wording of those other problems may have been quite different. So it's a matter of trying to find the equivalence to a problem whose solution you already know.
– David K
yesterday
I have done the first part and edit it.
– Mark
yesterday
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
yesterday
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
22 hours ago
Welcome to MSE. Usually people don’t answer to questions that show no effort on the questioner’s part because it makes the questioner appear to use this site as a homework help site, which it is not. Please edit your question to show more effort, such as any attempts.
– Kyky
yesterday
Welcome to MSE. Usually people don’t answer to questions that show no effort on the questioner’s part because it makes the questioner appear to use this site as a homework help site, which it is not. Please edit your question to show more effort, such as any attempts.
– Kyky
yesterday
Chances are that you've already seen problems equivalent to this one and have been given a method to solve them. But the wording of those other problems may have been quite different. So it's a matter of trying to find the equivalence to a problem whose solution you already know.
– David K
yesterday
Chances are that you've already seen problems equivalent to this one and have been given a method to solve them. But the wording of those other problems may have been quite different. So it's a matter of trying to find the equivalence to a problem whose solution you already know.
– David K
yesterday
I have done the first part and edit it.
– Mark
yesterday
I have done the first part and edit it.
– Mark
yesterday
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
yesterday
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
yesterday
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
22 hours ago
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
22 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
We can approach this in the following manner:
Let $a$ be the number of pennies chosen,
$b$ be the number of nickels chosen,
$c$ be the number of dimes chosen,
$d$ be the number of quarters chosen,
$e$ be the number of half-dollar coins chosen.
According to the given condition,
$a+b+c+d+e=11$
But $a,b,c,d,egeq 1$
So, substituting $p=a-1$, $q=b-1$, $r=c-1$, $s=d-1$, $t=e-1$,
We get ,
$p+q+r+s+t=6$ where $p,q,r,s,tgeq0$
The number of natural solutions will be ${(6+4) choose 4}$ i.e. $binom{10}{4} = 210$
We don't even have to worry about whether $a,b,c,d,e$ will exceed their limit of $10,10,10,9,8$ because we have chosen at least one from each pile which gives $max{a,b,c,d,e}$ possible $=7$.
answered 22 hours ago
Vaibhav
558
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We can approach this in the following manner:
Let $a$ be the number of pennies chosen,
$b$ be the number of nickels chosen,
$c$ be the number of dimes chosen,
$d$ be the number of quarters chosen,
$e$ be the number of half-dollar coins chosen.
According to the given condition,
$a+b+c+d+e=11$
But $a,b,c,d,egeq 1$
So, substituting $p=a-1$, $q=b-1$, $r=c-1$, $s=d-1$, $t=e-1$,
We get ,
$p+q+r+s+t=6$ where $p,q,r,s,tgeq0$
The number of natural solutions will be ${(6+4) choose 4}$ i.e. $binom{10}{4} = 210$
We don't even have to worry about whether $a,b,c,d,e$ will exceed their limit of $10,10,10,9,8$ because we have chosen at least one from each pile which gives $max{a,b,c,d,e}$ possible $=7$.
answered 22 hours ago
Vaibhav
558
add a comment |
up vote
1
down vote
We can approach this in the following manner:
Let $a$ be the number of pennies chosen,
$b$ be the number of nickels chosen,
$c$ be the number of dimes chosen,
$d$ be the number of quarters chosen,
$e$ be the number of half-dollar coins chosen.
According to the given condition,
$a+b+c+d+e=11$
But $a,b,c,d,egeq 1$
So, substituting $p=a-1$, $q=b-1$, $r=c-1$, $s=d-1$, $t=e-1$,
We get ,
$p+q+r+s+t=6$ where $p,q,r,s,tgeq0$
The number of natural solutions will be ${(6+4) choose 4}$ i.e. $binom{10}{4} = 210$
We don't even have to worry about whether $a,b,c,d,e$ will exceed their limit of $10,10,10,9,8$ because we have chosen at least one from each pile which gives $max{a,b,c,d,e}$ possible $=7$.
answered 22 hours ago
Vaibhav
558
add a comment |
up vote
1
down vote
up vote
1
down vote
We can approach this in the following manner:
Let $a$ be the number of pennies chosen,
$b$ be the number of nickels chosen,
$c$ be the number of dimes chosen,
$d$ be the number of quarters chosen,
$e$ be the number of half-dollar coins chosen.
According to the given condition,
$a+b+c+d+e=11$
But $a,b,c,d,egeq 1$
So, substituting $p=a-1$, $q=b-1$, $r=c-1$, $s=d-1$, $t=e-1$,
We get ,
$p+q+r+s+t=6$ where $p,q,r,s,tgeq0$
The number of natural solutions will be ${(6+4) choose 4}$ i.e. $binom{10}{4} = 210$
We don't even have to worry about whether $a,b,c,d,e$ will exceed their limit of $10,10,10,9,8$ because we have chosen at least one from each pile which gives $max{a,b,c,d,e}$ possible $=7$.
answered 22 hours ago
Vaibhav
558
We can approach this in the following manner:
Let $a$ be the number of pennies chosen,
$b$ be the number of nickels chosen,
$c$ be the number of dimes chosen,
$d$ be the number of quarters chosen,
$e$ be the number of half-dollar coins chosen.
According to the given condition,
$a+b+c+d+e=11$
But $a,b,c,d,egeq 1$
So, substituting $p=a-1$, $q=b-1$, $r=c-1$, $s=d-1$, $t=e-1$,
We get ,
$p+q+r+s+t=6$ where $p,q,r,s,tgeq0$
The number of natural solutions will be ${(6+4) choose 4}$ i.e. $binom{10}{4} = 210$
We don't even have to worry about whether $a,b,c,d,e$ will exceed their limit of $10,10,10,9,8$ because we have chosen at least one from each pile which gives $max{a,b,c,d,e}$ possible $=7$.
answered 22 hours ago
Vaibhav
558
answered 22 hours ago
Vaibhav
558
answered 22 hours ago
Vaibhav
558
answered 22 hours ago
Vaibhav
558
558
add a comment |
add a comment |
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Welcome to MSE. Usually people don’t answer to questions that show no effort on the questioner’s part because it makes the questioner appear to use this site as a homework help site, which it is not. Please edit your question to show more effort, such as any attempts.
– Kyky
yesterday
Chances are that you've already seen problems equivalent to this one and have been given a method to solve them. But the wording of those other problems may have been quite different. So it's a matter of trying to find the equivalence to a problem whose solution you already know.
– David K
yesterday
I have done the first part and edit it.
– Mark
yesterday
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
yesterday
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
22 hours ago