Drawing balls with a finite number of replacement












4














I have to solve this problem:



"Suppose a box contains $18$ balls numbered $1–6$, three balls with each number. When $4$ balls are drawn without replacement, how many outcomes are possible?". (The order does not matter).



I can't find a simple formula for it.
I've tried in this way and I don't know if it is right way:





A random outcome could or could not have the number $1$.
If it has it, the outcome could be $111$ plus a number $2le n le 6$, or $11$ plus two numbers or $1$ plus three numbers.




  • In the first case we have a total of ${{5}choose{1}} = 5 $ outcomes.

  • In the second case we have a total of ${{5}choose{2}} + 5 = 15$ outcomes.

  • In the last case we have a total of ${{5}choose{3}} + 5 +5times 4 = 35 $ outcomes.


Finally, if the outcomes does not have the number 1 we have a total of $ {{5}choose{4}} + 5times(4times 3 + 4) + 5times 4 + 5 = 110$.



So there are 165 possible outcomes.





Is it right? If yes, there is a simpler and much more elegant way to prove it?



Thanks










share|cite|improve this question





























    4














    I have to solve this problem:



    "Suppose a box contains $18$ balls numbered $1–6$, three balls with each number. When $4$ balls are drawn without replacement, how many outcomes are possible?". (The order does not matter).



    I can't find a simple formula for it.
    I've tried in this way and I don't know if it is right way:





    A random outcome could or could not have the number $1$.
    If it has it, the outcome could be $111$ plus a number $2le n le 6$, or $11$ plus two numbers or $1$ plus three numbers.




    • In the first case we have a total of ${{5}choose{1}} = 5 $ outcomes.

    • In the second case we have a total of ${{5}choose{2}} + 5 = 15$ outcomes.

    • In the last case we have a total of ${{5}choose{3}} + 5 +5times 4 = 35 $ outcomes.


    Finally, if the outcomes does not have the number 1 we have a total of $ {{5}choose{4}} + 5times(4times 3 + 4) + 5times 4 + 5 = 110$.



    So there are 165 possible outcomes.





    Is it right? If yes, there is a simpler and much more elegant way to prove it?



    Thanks










    share|cite|improve this question



























      4












      4








      4


      1





      I have to solve this problem:



      "Suppose a box contains $18$ balls numbered $1–6$, three balls with each number. When $4$ balls are drawn without replacement, how many outcomes are possible?". (The order does not matter).



      I can't find a simple formula for it.
      I've tried in this way and I don't know if it is right way:





      A random outcome could or could not have the number $1$.
      If it has it, the outcome could be $111$ plus a number $2le n le 6$, or $11$ plus two numbers or $1$ plus three numbers.




      • In the first case we have a total of ${{5}choose{1}} = 5 $ outcomes.

      • In the second case we have a total of ${{5}choose{2}} + 5 = 15$ outcomes.

      • In the last case we have a total of ${{5}choose{3}} + 5 +5times 4 = 35 $ outcomes.


      Finally, if the outcomes does not have the number 1 we have a total of $ {{5}choose{4}} + 5times(4times 3 + 4) + 5times 4 + 5 = 110$.



      So there are 165 possible outcomes.





      Is it right? If yes, there is a simpler and much more elegant way to prove it?



      Thanks










      share|cite|improve this question















      I have to solve this problem:



      "Suppose a box contains $18$ balls numbered $1–6$, three balls with each number. When $4$ balls are drawn without replacement, how many outcomes are possible?". (The order does not matter).



      I can't find a simple formula for it.
      I've tried in this way and I don't know if it is right way:





      A random outcome could or could not have the number $1$.
      If it has it, the outcome could be $111$ plus a number $2le n le 6$, or $11$ plus two numbers or $1$ plus three numbers.




      • In the first case we have a total of ${{5}choose{1}} = 5 $ outcomes.

      • In the second case we have a total of ${{5}choose{2}} + 5 = 15$ outcomes.

      • In the last case we have a total of ${{5}choose{3}} + 5 +5times 4 = 35 $ outcomes.


      Finally, if the outcomes does not have the number 1 we have a total of $ {{5}choose{4}} + 5times(4times 3 + 4) + 5times 4 + 5 = 110$.



      So there are 165 possible outcomes.





      Is it right? If yes, there is a simpler and much more elegant way to prove it?



      Thanks







      probability combinatorics permutations combinations






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      edited Dec 20 '18 at 17:14









      J.G.

      23k22137




      23k22137










      asked Dec 20 '18 at 14:21









      Alex

      32319




      32319






















          4 Answers
          4






          active

          oldest

          votes


















          8














          Method I. (case by case)



          Case I: all numbers are distinct. Then there are $binom 64 =15$.



          Case II: one pair, the other two distinct. Then there are $6$ ways to choose the rank of the pair, and $binom 52=10$ ways to choose the odd two. Thus $6times 10=60$.



          Case III: two pairs. There are $binom 62=15$.



          Case IV: one triple. There are $6$ ways to choose the rank of the triple and $5$ ways to choose the odd man out. Thus $30$.



          So: $$15+60+15+30=120$$



          Method II. (Stars and Bars)



          If you ignore the fact that we only have three of each number, then we are just asking for the number of $6-$tuples of non-negative integers that sum to $4$. Stars and Bars then tells us that the answer would be $binom {4+6-1}4=binom 94=126$. Of course, this also counts the six cases in which all the balls show the same number. As that is impossible, we must remove those cases, so the answer would then be $$126-6=120$$






          share|cite|improve this answer































            3














            First let's solve it for $24$ balls numbered $1-6$, $4$ balls of each number.



            Then to be found is the number of sums $a_1+a_2+a_3+a_4+a_5+a_6=4$ where the $a_i$ are nonnegative integers.



            Here $a_i$ corresponds with the number of balls that are drawn and carry number $i$.



            With stars and bars we find $binom{9}{5}=126$ possibilities.



            Now we must subtract the number of "possibilities" that actually were not possibilities because $4$ balls having the same number were drawn.



            Then we end up with $$binom95-6=120$$ possibilities.






            share|cite|improve this answer





























              2














              One more way is to use a generating function. Consider



              $$F(x) = (1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)cdots(1+x+x^2+x^3) = (1+x+x^2+x^3)^6$$



              Looking at the first $(1 + x+ x^2 + x^3)$ term, we can think of the exponent of $x$ as representing the number of "1" balls we choose (i.e. $1=x^0 rightarrow 0, x = x^1rightarrow 1, x^2rightarrow 2, x^3rightarrow 3$). Correspondingly, the second term represents the number of "2" balls we choose, etc. Since the exponents in each term range from 0 to 3, we are restricted to choosing at most 3 of any type of ball. The answer to your question is then given by the coefficient of $x^4$ (since we are choosing a total of 4 balls) in the expansion of $F(x)$. A computer can easily confirm that the coefficient is 120, as given by the other answers.



              This method is nice because it can be generalized to more complicated conditions fairly easily. For example, if you are only interested in the number of drawings where an even number of "1" balls are present, you can simply change the first term to be $(1 + x^2)$ (note the even exponents) and find the coefficient again.






              share|cite|improve this answer





























                1














                Just for fun, and because it sometimes helps to see the things we are counting:



                1112 1113 1114 1115 1116 1122 1123 1124 1125 1126 1133 1134 
                1135 1136 1144 1145 1146 1155 1156 1166 1222 1223 1224 1225
                1226 1233 1234 1235 1236 1244 1245 1246 1255 1256 1266 1333
                1334 1335 1336 1344 1345 1346 1355 1356 1366 1444 1445 1446
                1455 1456 1466 1555 1556 1566 1666 2223 2224 2225 2226 2233
                2234 2235 2236 2244 2245 2246 2255 2256 2266 2333 2334 2335
                2336 2344 2345 2346 2355 2356 2366 2444 2445 2446 2455 2456
                2466 2555 2556 2566 2666 3334 3335 3336 3344 3345 3346 3355
                3356 3366 3444 3445 3446 3455 3456 3466 3555 3556 3566 3666
                4445 4446 4455 4456 4466 4555 4556 4566 4666 5556 5566 5666





                share|cite|improve this answer





















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                  4 Answers
                  4






                  active

                  oldest

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                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  8














                  Method I. (case by case)



                  Case I: all numbers are distinct. Then there are $binom 64 =15$.



                  Case II: one pair, the other two distinct. Then there are $6$ ways to choose the rank of the pair, and $binom 52=10$ ways to choose the odd two. Thus $6times 10=60$.



                  Case III: two pairs. There are $binom 62=15$.



                  Case IV: one triple. There are $6$ ways to choose the rank of the triple and $5$ ways to choose the odd man out. Thus $30$.



                  So: $$15+60+15+30=120$$



                  Method II. (Stars and Bars)



                  If you ignore the fact that we only have three of each number, then we are just asking for the number of $6-$tuples of non-negative integers that sum to $4$. Stars and Bars then tells us that the answer would be $binom {4+6-1}4=binom 94=126$. Of course, this also counts the six cases in which all the balls show the same number. As that is impossible, we must remove those cases, so the answer would then be $$126-6=120$$






                  share|cite|improve this answer




























                    8














                    Method I. (case by case)



                    Case I: all numbers are distinct. Then there are $binom 64 =15$.



                    Case II: one pair, the other two distinct. Then there are $6$ ways to choose the rank of the pair, and $binom 52=10$ ways to choose the odd two. Thus $6times 10=60$.



                    Case III: two pairs. There are $binom 62=15$.



                    Case IV: one triple. There are $6$ ways to choose the rank of the triple and $5$ ways to choose the odd man out. Thus $30$.



                    So: $$15+60+15+30=120$$



                    Method II. (Stars and Bars)



                    If you ignore the fact that we only have three of each number, then we are just asking for the number of $6-$tuples of non-negative integers that sum to $4$. Stars and Bars then tells us that the answer would be $binom {4+6-1}4=binom 94=126$. Of course, this also counts the six cases in which all the balls show the same number. As that is impossible, we must remove those cases, so the answer would then be $$126-6=120$$






                    share|cite|improve this answer


























                      8












                      8








                      8






                      Method I. (case by case)



                      Case I: all numbers are distinct. Then there are $binom 64 =15$.



                      Case II: one pair, the other two distinct. Then there are $6$ ways to choose the rank of the pair, and $binom 52=10$ ways to choose the odd two. Thus $6times 10=60$.



                      Case III: two pairs. There are $binom 62=15$.



                      Case IV: one triple. There are $6$ ways to choose the rank of the triple and $5$ ways to choose the odd man out. Thus $30$.



                      So: $$15+60+15+30=120$$



                      Method II. (Stars and Bars)



                      If you ignore the fact that we only have three of each number, then we are just asking for the number of $6-$tuples of non-negative integers that sum to $4$. Stars and Bars then tells us that the answer would be $binom {4+6-1}4=binom 94=126$. Of course, this also counts the six cases in which all the balls show the same number. As that is impossible, we must remove those cases, so the answer would then be $$126-6=120$$






                      share|cite|improve this answer














                      Method I. (case by case)



                      Case I: all numbers are distinct. Then there are $binom 64 =15$.



                      Case II: one pair, the other two distinct. Then there are $6$ ways to choose the rank of the pair, and $binom 52=10$ ways to choose the odd two. Thus $6times 10=60$.



                      Case III: two pairs. There are $binom 62=15$.



                      Case IV: one triple. There are $6$ ways to choose the rank of the triple and $5$ ways to choose the odd man out. Thus $30$.



                      So: $$15+60+15+30=120$$



                      Method II. (Stars and Bars)



                      If you ignore the fact that we only have three of each number, then we are just asking for the number of $6-$tuples of non-negative integers that sum to $4$. Stars and Bars then tells us that the answer would be $binom {4+6-1}4=binom 94=126$. Of course, this also counts the six cases in which all the balls show the same number. As that is impossible, we must remove those cases, so the answer would then be $$126-6=120$$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 20 '18 at 15:55









                      Connor Harris

                      4,303723




                      4,303723










                      answered Dec 20 '18 at 14:53









                      lulu

                      39.2k24677




                      39.2k24677























                          3














                          First let's solve it for $24$ balls numbered $1-6$, $4$ balls of each number.



                          Then to be found is the number of sums $a_1+a_2+a_3+a_4+a_5+a_6=4$ where the $a_i$ are nonnegative integers.



                          Here $a_i$ corresponds with the number of balls that are drawn and carry number $i$.



                          With stars and bars we find $binom{9}{5}=126$ possibilities.



                          Now we must subtract the number of "possibilities" that actually were not possibilities because $4$ balls having the same number were drawn.



                          Then we end up with $$binom95-6=120$$ possibilities.






                          share|cite|improve this answer


























                            3














                            First let's solve it for $24$ balls numbered $1-6$, $4$ balls of each number.



                            Then to be found is the number of sums $a_1+a_2+a_3+a_4+a_5+a_6=4$ where the $a_i$ are nonnegative integers.



                            Here $a_i$ corresponds with the number of balls that are drawn and carry number $i$.



                            With stars and bars we find $binom{9}{5}=126$ possibilities.



                            Now we must subtract the number of "possibilities" that actually were not possibilities because $4$ balls having the same number were drawn.



                            Then we end up with $$binom95-6=120$$ possibilities.






                            share|cite|improve this answer
























                              3












                              3








                              3






                              First let's solve it for $24$ balls numbered $1-6$, $4$ balls of each number.



                              Then to be found is the number of sums $a_1+a_2+a_3+a_4+a_5+a_6=4$ where the $a_i$ are nonnegative integers.



                              Here $a_i$ corresponds with the number of balls that are drawn and carry number $i$.



                              With stars and bars we find $binom{9}{5}=126$ possibilities.



                              Now we must subtract the number of "possibilities" that actually were not possibilities because $4$ balls having the same number were drawn.



                              Then we end up with $$binom95-6=120$$ possibilities.






                              share|cite|improve this answer












                              First let's solve it for $24$ balls numbered $1-6$, $4$ balls of each number.



                              Then to be found is the number of sums $a_1+a_2+a_3+a_4+a_5+a_6=4$ where the $a_i$ are nonnegative integers.



                              Here $a_i$ corresponds with the number of balls that are drawn and carry number $i$.



                              With stars and bars we find $binom{9}{5}=126$ possibilities.



                              Now we must subtract the number of "possibilities" that actually were not possibilities because $4$ balls having the same number were drawn.



                              Then we end up with $$binom95-6=120$$ possibilities.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 20 '18 at 14:50









                              drhab

                              98k544129




                              98k544129























                                  2














                                  One more way is to use a generating function. Consider



                                  $$F(x) = (1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)cdots(1+x+x^2+x^3) = (1+x+x^2+x^3)^6$$



                                  Looking at the first $(1 + x+ x^2 + x^3)$ term, we can think of the exponent of $x$ as representing the number of "1" balls we choose (i.e. $1=x^0 rightarrow 0, x = x^1rightarrow 1, x^2rightarrow 2, x^3rightarrow 3$). Correspondingly, the second term represents the number of "2" balls we choose, etc. Since the exponents in each term range from 0 to 3, we are restricted to choosing at most 3 of any type of ball. The answer to your question is then given by the coefficient of $x^4$ (since we are choosing a total of 4 balls) in the expansion of $F(x)$. A computer can easily confirm that the coefficient is 120, as given by the other answers.



                                  This method is nice because it can be generalized to more complicated conditions fairly easily. For example, if you are only interested in the number of drawings where an even number of "1" balls are present, you can simply change the first term to be $(1 + x^2)$ (note the even exponents) and find the coefficient again.






                                  share|cite|improve this answer


























                                    2














                                    One more way is to use a generating function. Consider



                                    $$F(x) = (1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)cdots(1+x+x^2+x^3) = (1+x+x^2+x^3)^6$$



                                    Looking at the first $(1 + x+ x^2 + x^3)$ term, we can think of the exponent of $x$ as representing the number of "1" balls we choose (i.e. $1=x^0 rightarrow 0, x = x^1rightarrow 1, x^2rightarrow 2, x^3rightarrow 3$). Correspondingly, the second term represents the number of "2" balls we choose, etc. Since the exponents in each term range from 0 to 3, we are restricted to choosing at most 3 of any type of ball. The answer to your question is then given by the coefficient of $x^4$ (since we are choosing a total of 4 balls) in the expansion of $F(x)$. A computer can easily confirm that the coefficient is 120, as given by the other answers.



                                    This method is nice because it can be generalized to more complicated conditions fairly easily. For example, if you are only interested in the number of drawings where an even number of "1" balls are present, you can simply change the first term to be $(1 + x^2)$ (note the even exponents) and find the coefficient again.






                                    share|cite|improve this answer
























                                      2












                                      2








                                      2






                                      One more way is to use a generating function. Consider



                                      $$F(x) = (1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)cdots(1+x+x^2+x^3) = (1+x+x^2+x^3)^6$$



                                      Looking at the first $(1 + x+ x^2 + x^3)$ term, we can think of the exponent of $x$ as representing the number of "1" balls we choose (i.e. $1=x^0 rightarrow 0, x = x^1rightarrow 1, x^2rightarrow 2, x^3rightarrow 3$). Correspondingly, the second term represents the number of "2" balls we choose, etc. Since the exponents in each term range from 0 to 3, we are restricted to choosing at most 3 of any type of ball. The answer to your question is then given by the coefficient of $x^4$ (since we are choosing a total of 4 balls) in the expansion of $F(x)$. A computer can easily confirm that the coefficient is 120, as given by the other answers.



                                      This method is nice because it can be generalized to more complicated conditions fairly easily. For example, if you are only interested in the number of drawings where an even number of "1" balls are present, you can simply change the first term to be $(1 + x^2)$ (note the even exponents) and find the coefficient again.






                                      share|cite|improve this answer












                                      One more way is to use a generating function. Consider



                                      $$F(x) = (1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)cdots(1+x+x^2+x^3) = (1+x+x^2+x^3)^6$$



                                      Looking at the first $(1 + x+ x^2 + x^3)$ term, we can think of the exponent of $x$ as representing the number of "1" balls we choose (i.e. $1=x^0 rightarrow 0, x = x^1rightarrow 1, x^2rightarrow 2, x^3rightarrow 3$). Correspondingly, the second term represents the number of "2" balls we choose, etc. Since the exponents in each term range from 0 to 3, we are restricted to choosing at most 3 of any type of ball. The answer to your question is then given by the coefficient of $x^4$ (since we are choosing a total of 4 balls) in the expansion of $F(x)$. A computer can easily confirm that the coefficient is 120, as given by the other answers.



                                      This method is nice because it can be generalized to more complicated conditions fairly easily. For example, if you are only interested in the number of drawings where an even number of "1" balls are present, you can simply change the first term to be $(1 + x^2)$ (note the even exponents) and find the coefficient again.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 20 '18 at 21:04









                                      Alex

                                      1777




                                      1777























                                          1














                                          Just for fun, and because it sometimes helps to see the things we are counting:



                                          1112 1113 1114 1115 1116 1122 1123 1124 1125 1126 1133 1134 
                                          1135 1136 1144 1145 1146 1155 1156 1166 1222 1223 1224 1225
                                          1226 1233 1234 1235 1236 1244 1245 1246 1255 1256 1266 1333
                                          1334 1335 1336 1344 1345 1346 1355 1356 1366 1444 1445 1446
                                          1455 1456 1466 1555 1556 1566 1666 2223 2224 2225 2226 2233
                                          2234 2235 2236 2244 2245 2246 2255 2256 2266 2333 2334 2335
                                          2336 2344 2345 2346 2355 2356 2366 2444 2445 2446 2455 2456
                                          2466 2555 2556 2566 2666 3334 3335 3336 3344 3345 3346 3355
                                          3356 3366 3444 3445 3446 3455 3456 3466 3555 3556 3566 3666
                                          4445 4446 4455 4456 4466 4555 4556 4566 4666 5556 5566 5666





                                          share|cite|improve this answer


























                                            1














                                            Just for fun, and because it sometimes helps to see the things we are counting:



                                            1112 1113 1114 1115 1116 1122 1123 1124 1125 1126 1133 1134 
                                            1135 1136 1144 1145 1146 1155 1156 1166 1222 1223 1224 1225
                                            1226 1233 1234 1235 1236 1244 1245 1246 1255 1256 1266 1333
                                            1334 1335 1336 1344 1345 1346 1355 1356 1366 1444 1445 1446
                                            1455 1456 1466 1555 1556 1566 1666 2223 2224 2225 2226 2233
                                            2234 2235 2236 2244 2245 2246 2255 2256 2266 2333 2334 2335
                                            2336 2344 2345 2346 2355 2356 2366 2444 2445 2446 2455 2456
                                            2466 2555 2556 2566 2666 3334 3335 3336 3344 3345 3346 3355
                                            3356 3366 3444 3445 3446 3455 3456 3466 3555 3556 3566 3666
                                            4445 4446 4455 4456 4466 4555 4556 4566 4666 5556 5566 5666





                                            share|cite|improve this answer
























                                              1












                                              1








                                              1






                                              Just for fun, and because it sometimes helps to see the things we are counting:



                                              1112 1113 1114 1115 1116 1122 1123 1124 1125 1126 1133 1134 
                                              1135 1136 1144 1145 1146 1155 1156 1166 1222 1223 1224 1225
                                              1226 1233 1234 1235 1236 1244 1245 1246 1255 1256 1266 1333
                                              1334 1335 1336 1344 1345 1346 1355 1356 1366 1444 1445 1446
                                              1455 1456 1466 1555 1556 1566 1666 2223 2224 2225 2226 2233
                                              2234 2235 2236 2244 2245 2246 2255 2256 2266 2333 2334 2335
                                              2336 2344 2345 2346 2355 2356 2366 2444 2445 2446 2455 2456
                                              2466 2555 2556 2566 2666 3334 3335 3336 3344 3345 3346 3355
                                              3356 3366 3444 3445 3446 3455 3456 3466 3555 3556 3566 3666
                                              4445 4446 4455 4456 4466 4555 4556 4566 4666 5556 5566 5666





                                              share|cite|improve this answer












                                              Just for fun, and because it sometimes helps to see the things we are counting:



                                              1112 1113 1114 1115 1116 1122 1123 1124 1125 1126 1133 1134 
                                              1135 1136 1144 1145 1146 1155 1156 1166 1222 1223 1224 1225
                                              1226 1233 1234 1235 1236 1244 1245 1246 1255 1256 1266 1333
                                              1334 1335 1336 1344 1345 1346 1355 1356 1366 1444 1445 1446
                                              1455 1456 1466 1555 1556 1566 1666 2223 2224 2225 2226 2233
                                              2234 2235 2236 2244 2245 2246 2255 2256 2266 2333 2334 2335
                                              2336 2344 2345 2346 2355 2356 2366 2444 2445 2446 2455 2456
                                              2466 2555 2556 2566 2666 3334 3335 3336 3344 3345 3346 3355
                                              3356 3366 3444 3445 3446 3455 3456 3466 3555 3556 3566 3666
                                              4445 4446 4455 4456 4466 4555 4556 4566 4666 5556 5566 5666






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                                              answered Dec 20 '18 at 15:25









                                              Richard Ambler

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