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 Nov 18 at 13:46
Mark
11
add a comment |
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 Nov 18 at 13:46
Mark
11
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
Nov 18 at 13:55
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
Nov 18 at 14:58
I have done the first part and edit it.
– Mark
Nov 19 at 4:22
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
Nov 19 at 4:25
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
Nov 19 at 7:30
add a comment |
up vote
0
down vote
favorite
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 Nov 18 at 13:46
Mark
11
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 Nov 18 at 13:46
Mark
11
asked Nov 18 at 13:46
Mark
11
edited Nov 19 at 7:06
asked Nov 18 at 13:46
Mark
11
asked Nov 18 at 13:46
Mark
11
asked Nov 18 at 13:46
Mark
11
11
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
Nov 18 at 13:55
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
Nov 18 at 14:58
I have done the first part and edit it.
– Mark
Nov 19 at 4:22
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
Nov 19 at 4:25
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
Nov 19 at 7:30
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
Nov 18 at 13:55
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
Nov 18 at 14:58
I have done the first part and edit it.
– Mark
Nov 19 at 4:22
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
Nov 19 at 4:25
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
Nov 19 at 7:30
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
Nov 18 at 13:55
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
Nov 18 at 13:55
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
Nov 18 at 14:58
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
Nov 18 at 14:58
I have done the first part and edit it.
– Mark
Nov 19 at 4:22
I have done the first part and edit it.
– Mark
Nov 19 at 4:22
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
Nov 19 at 4:25
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
Nov 19 at 4:25
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
Nov 19 at 7:30
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
Nov 19 at 7:30
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 Nov 19 at 7:28
Vaibhav
588
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 Nov 19 at 7:28
Vaibhav
588
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 Nov 19 at 7:28
Vaibhav
588
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 Nov 19 at 7:28
Vaibhav
588
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 Nov 19 at 7:28
Vaibhav
588
answered Nov 19 at 7:28
Vaibhav
588
answered Nov 19 at 7:28
Vaibhav
588
answered Nov 19 at 7:28
Vaibhav
588
588
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
Nov 18 at 13:55
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
Nov 18 at 14:58
I have done the first part and edit it.
– Mark
Nov 19 at 4:22
What is $15$ in $15 C11$? $11$ are the coin that you choose, but what is $15$?
– Jimmy R.
Nov 19 at 4:25
@user525966 That approach is wrong because $binom{42}{6}$ represents the situation when all the coins are unique.
– Vaibhav
Nov 19 at 7:30