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0

$begingroup$

Alice has 10 balls (all different). First, she splits them into two piles; then she picks one of the piles with at least two elements, and then splits it into two; she repeats this until each pile has only one element. Find the number of different ways in which she can carry out this procedure.

I know that this is a classic question and I checked the answer that was responded in this question.
I just need an explanation.

We have 10 balls and we want to divide them into 2 groups until we are left with one ball. So my first intuition was $10choose9*$$10choose8*$$10choose7*$$10choose6*$$10choose5$ then for each of the possible groups I wanted to find every possible combination of groups and realized that this is not very smart. How can I know that this question can be reversible and reach to the answer: $10choose2*$$9choose2*$...$3choose2*$$2choose2$, also If I spent all my time and did my intuition would I get the same result?

combinatorics combinations

share|cite|improve this question

asked Mar 13 at 18:01

70pr4k70pr4k

678

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  isn't the explanation already given in the linked post sufficient?
  $endgroup$
  – G Cab
  Mar 13 at 18:23
  

  
  
  
  

add a comment | 

0

$begingroup$

Alice has 10 balls (all different). First, she splits them into two piles; then she picks one of the piles with at least two elements, and then splits it into two; she repeats this until each pile has only one element. Find the number of different ways in which she can carry out this procedure.

I know that this is a classic question and I checked the answer that was responded in this question.
I just need an explanation.

We have 10 balls and we want to divide them into 2 groups until we are left with one ball. So my first intuition was $10choose9*$$10choose8*$$10choose7*$$10choose6*$$10choose5$ then for each of the possible groups I wanted to find every possible combination of groups and realized that this is not very smart. How can I know that this question can be reversible and reach to the answer: $10choose2*$$9choose2*$...$3choose2*$$2choose2$, also If I spent all my time and did my intuition would I get the same result?

combinatorics combinations

share|cite|improve this question

asked Mar 13 at 18:01

70pr4k70pr4k

678

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  isn't the explanation already given in the linked post sufficient?
  $endgroup$
  – G Cab
  Mar 13 at 18:23
  

  
  
  
  

add a comment | 

$begingroup$

Alice has 10 balls (all different). First, she splits them into two piles; then she picks one of the piles with at least two elements, and then splits it into two; she repeats this until each pile has only one element. Find the number of different ways in which she can carry out this procedure.

I know that this is a classic question and I checked the answer that was responded in this question.
I just need an explanation.

We have 10 balls and we want to divide them into 2 groups until we are left with one ball. So my first intuition was $10choose9*$$10choose8*$$10choose7*$$10choose6*$$10choose5$ then for each of the possible groups I wanted to find every possible combination of groups and realized that this is not very smart. How can I know that this question can be reversible and reach to the answer: $10choose2*$$9choose2*$...$3choose2*$$2choose2$, also If I spent all my time and did my intuition would I get the same result?

combinatorics combinations

share|cite|improve this question

asked Mar 13 at 18:01

70pr4k70pr4k

678

$endgroup$

share|cite|improve this question

asked Mar 13 at 18:01

70pr4k70pr4k

678

share|cite|improve this question

asked Mar 13 at 18:01

70pr4k70pr4k

678

asked Mar 13 at 18:01

70pr4k70pr4k

678

asked Mar 13 at 18:01

70pr4k70pr4k

678