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







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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















up vote
0
down 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







share|cite|improve this question









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













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







share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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Mark is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 22 hours ago





















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asked yesterday









Mark

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11




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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


















  • 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










1 Answer
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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 $=7$.






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    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 $=7$.






    share|cite|improve this answer

























      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 $=7$.






      share|cite|improve this answer























        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 $=7$.






        share|cite|improve this answer












        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 $=7$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 22 hours ago









        Vaibhav

        558




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