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Choosing $n$ and $k$ in combinations

Use of a null option in a combination with repetition problemWhy is my answer to this multichoose counting problem wrong?Combinations with RepetitionCombinations with Repetition Versus … what is this?How would you work out these combinations?Number of combinations of ice-cream with limited quantity.Combinations with ice creamProbability when choosing over more combinationsAn ice cream parlor has 28 different ice cream flavors. How many different ways are there to choose 6 scoops of ice cream if at leastCounting problem.How many different bowls can be made if:

0

1

$begingroup$

We have $5$ kinds of ice cream with each kind having $10$ pieces. For orders containing $10$ icecreams, how many different combinations are possible? The answer should be $1001$

= we are choosing from $5$ kinds of ice cream so $n = 5$ and $k = 10$

using formula

$binomn + k - 1 n = binom n + k - 1 k - 1 $ so
$binom 5 + 10 - 1 5 $ but this equals to $2002$. But if I choose $k = 5$ and $n = 10$, it will results in $1001$, which is the correct answer$ldots$ So why do we have to choose $n$ as $10$ and not $5$? We are choosing from $5$ different kinds so it should be $5$. Or am i mistaken?

combinatorics combinations

share|cite|improve this question

edited Mar 17 at 11:11

Rócherz

3,0013821

asked Nov 20 '16 at 0:00

J.ddJ.dd

182

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Make sure that you understand what the letters used in the formula are meant to represent. $binombins+balls-1bins-1$. Here, the flavors (five of them) act as the bins, and the number of scoops act as the balls.
  $endgroup$
  – JMoravitz
  Nov 20 '16 at 0:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I thought i do , using example when we have 6 red 6 blue and 6 green balls and we have to find the number of ways how we can take 6 balls we choose n=3
  $endgroup$
  – J.dd
  Nov 20 '16 at 0:03
  

  
  
  
  

add a comment | 

0

1

$begingroup$

We have $5$ kinds of ice cream with each kind having $10$ pieces. For orders containing $10$ icecreams, how many different combinations are possible? The answer should be $1001$

= we are choosing from $5$ kinds of ice cream so $n = 5$ and $k = 10$

using formula

$binomn + k - 1 n = binom n + k - 1 k - 1 $ so
$binom 5 + 10 - 1 5 $ but this equals to $2002$. But if I choose $k = 5$ and $n = 10$, it will results in $1001$, which is the correct answer$ldots$ So why do we have to choose $n$ as $10$ and not $5$? We are choosing from $5$ different kinds so it should be $5$. Or am i mistaken?

combinatorics combinations

share|cite|improve this question

edited Mar 17 at 11:11

Rócherz

3,0013821

asked Nov 20 '16 at 0:00

J.ddJ.dd

182

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Make sure that you understand what the letters used in the formula are meant to represent. $binombins+balls-1bins-1$. Here, the flavors (five of them) act as the bins, and the number of scoops act as the balls.
  $endgroup$
  – JMoravitz
  Nov 20 '16 at 0:02
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I thought i do , using example when we have 6 red 6 blue and 6 green balls and we have to find the number of ways how we can take 6 balls we choose n=3
  $endgroup$
  – J.dd
  Nov 20 '16 at 0:03
  

  
  
  
  

add a comment | 

$begingroup$

We have $5$ kinds of ice cream with each kind having $10$ pieces. For orders containing $10$ icecreams, how many different combinations are possible? The answer should be $1001$

= we are choosing from $5$ kinds of ice cream so $n = 5$ and $k = 10$

using formula

$binomn + k - 1 n = binom n + k - 1 k - 1 $ so
$binom 5 + 10 - 1 5 $ but this equals to $2002$. But if I choose $k = 5$ and $n = 10$, it will results in $1001$, which is the correct answer$ldots$ So why do we have to choose $n$ as $10$ and not $5$? We are choosing from $5$ different kinds so it should be $5$. Or am i mistaken?

combinatorics combinations

share|cite|improve this question

edited Mar 17 at 11:11

Rócherz

3,0013821

asked Nov 20 '16 at 0:00

J.ddJ.dd

182

$endgroup$

share|cite|improve this question

edited Mar 17 at 11:11

Rócherz

3,0013821

asked Nov 20 '16 at 0:00

J.ddJ.dd

182

share|cite|improve this question

edited Mar 17 at 11:11

Rócherz

3,0013821

asked Nov 20 '16 at 0:00

J.ddJ.dd

182

edited Mar 17 at 11:11

Rócherz

3,0013821

edited Mar 17 at 11:11

Rócherz

3,0013821

asked Nov 20 '16 at 0:00

J.ddJ.dd

182

asked Nov 20 '16 at 0:00

J.ddJ.dd

182

1 Answer
1

active

oldest

votes

0

$begingroup$

Ordering 10 icecreams out of 5 kinds, is nothing else as putting 10 balls into 5 bins. So $k=5$ and $n=10$ in full agreement with the correct answer.

Possibly you misinterpret the meaning of $n $ and $k $ in the expression
$$binom n+k-1n. $$

The correct meaning can be easily memorized using stars and bars approach.

share|cite|improve this answer

edited Mar 17 at 12:28

answered Mar 17 at 12:07

useruser

5,93011031

$endgroup$

add a comment | 

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0

$begingroup$

Ordering 10 icecreams out of 5 kinds, is nothing else as putting 10 balls into 5 bins. So $k=5$ and $n=10$ in full agreement with the correct answer.

Possibly you misinterpret the meaning of $n $ and $k $ in the expression
$$binom n+k-1n. $$

The correct meaning can be easily memorized using stars and bars approach.

share|cite|improve this answer

edited Mar 17 at 12:28

answered Mar 17 at 12:07

useruser

5,93011031

$endgroup$

add a comment | 

0

$begingroup$

Ordering 10 icecreams out of 5 kinds, is nothing else as putting 10 balls into 5 bins. So $k=5$ and $n=10$ in full agreement with the correct answer.

Possibly you misinterpret the meaning of $n $ and $k $ in the expression
$$binom n+k-1n. $$

The correct meaning can be easily memorized using stars and bars approach.

share|cite|improve this answer

edited Mar 17 at 12:28

answered Mar 17 at 12:07

useruser

5,93011031

$endgroup$

add a comment | 

$begingroup$

Ordering 10 icecreams out of 5 kinds, is nothing else as putting 10 balls into 5 bins. So $k=5$ and $n=10$ in full agreement with the correct answer.

Possibly you misinterpret the meaning of $n $ and $k $ in the expression
$$binom n+k-1n. $$

The correct meaning can be easily memorized using stars and bars approach.

share|cite|improve this answer

edited Mar 17 at 12:28

answered Mar 17 at 12:07

useruser

5,93011031

$endgroup$

share|cite|improve this answer

edited Mar 17 at 12:28

answered Mar 17 at 12:07

useruser

5,93011031

answered Mar 17 at 12:07

useruser

5,93011031

answered Mar 17 at 12:07

useruser

5,93011031