Find number of occurrences of $n$ in a combination

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list $= 1, 2, 3, 4, 5$;

The above list has $5$ numbers.
The number of combinations when we group them into $3$ is $10$.

The list if combinations are

$123, 124, 125, 134, 135, 145, 234, 235, 245, 345$

I need to calculate the number of occurrences of an item in the list.
When $n = 5$, the number of occurrences of $n$ in the list is $6$.

Is there any formula to find the number of occurrences?

combinatorics combinations

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edited Jul 17 at 6:58

N. F. Taussig

38.3k93053

asked Jul 17 at 6:44

Karthik Sankar

13

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The combinations with $5$ correspond to two-number combinations from $1,2,3,4,6$.
  – Lord Shark the Unknown
  Jul 17 at 6:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So what you're asking is: When choosing $k$ items from a set of length $n$, how many of them include the item number $l$, so that the order does not matter? You can think about it like this: since the order does not matter, you can first choose the element that you are listing the occurrance of. After that, you are choosing $k-1$ items from a list of $n-1$.
  – Matti P.
  Jul 17 at 6:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @LordSharktheUnknown You mean from the set $1,2,3,4$, and indeed $4 choose 2 = 6$
  – Brevan Ellefsen
  Jul 17 at 6:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MattiP. Sorry for the inconvenience. I can't understand what you are pointing to n. I've updated the question. Can u pls read again!
  – Karthik Sankar
  Jul 17 at 6:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, I see that our notations are contradicting. What you call $n$ in the question is my $l$. What I mean by $n$ is the length of the list, which also has the fixed value $n=5$ in this case.
  – Matti P.
  Jul 17 at 7:38
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

list $= 1, 2, 3, 4, 5$;

The above list has $5$ numbers.
The number of combinations when we group them into $3$ is $10$.

The list if combinations are

$123, 124, 125, 134, 135, 145, 234, 235, 245, 345$

I need to calculate the number of occurrences of an item in the list.
When $n = 5$, the number of occurrences of $n$ in the list is $6$.

Is there any formula to find the number of occurrences?

combinatorics combinations

share|cite|improve this question

edited Jul 17 at 6:58

N. F. Taussig

38.3k93053

asked Jul 17 at 6:44

Karthik Sankar

13

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The combinations with $5$ correspond to two-number combinations from $1,2,3,4,6$.
  – Lord Shark the Unknown
  Jul 17 at 6:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So what you're asking is: When choosing $k$ items from a set of length $n$, how many of them include the item number $l$, so that the order does not matter? You can think about it like this: since the order does not matter, you can first choose the element that you are listing the occurrance of. After that, you are choosing $k-1$ items from a list of $n-1$.
  – Matti P.
  Jul 17 at 6:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @LordSharktheUnknown You mean from the set $1,2,3,4$, and indeed $4 choose 2 = 6$
  – Brevan Ellefsen
  Jul 17 at 6:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MattiP. Sorry for the inconvenience. I can't understand what you are pointing to n. I've updated the question. Can u pls read again!
  – Karthik Sankar
  Jul 17 at 6:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, I see that our notations are contradicting. What you call $n$ in the question is my $l$. What I mean by $n$ is the length of the list, which also has the fixed value $n=5$ in this case.
  – Matti P.
  Jul 17 at 7:38
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

list $= 1, 2, 3, 4, 5$;

The above list has $5$ numbers.
The number of combinations when we group them into $3$ is $10$.

The list if combinations are

$123, 124, 125, 134, 135, 145, 234, 235, 245, 345$

I need to calculate the number of occurrences of an item in the list.
When $n = 5$, the number of occurrences of $n$ in the list is $6$.

Is there any formula to find the number of occurrences?

combinatorics combinations

share|cite|improve this question

edited Jul 17 at 6:58

N. F. Taussig

38.3k93053

asked Jul 17 at 6:44

Karthik Sankar

13

list $= 1, 2, 3, 4, 5$;

The above list has $5$ numbers.
The number of combinations when we group them into $3$ is $10$.

The list if combinations are

$123, 124, 125, 134, 135, 145, 234, 235, 245, 345$

I need to calculate the number of occurrences of an item in the list.
When $n = 5$, the number of occurrences of $n$ in the list is $6$.

Is there any formula to find the number of occurrences?

combinatorics combinations

share|cite|improve this question

edited Jul 17 at 6:58

N. F. Taussig

38.3k93053

asked Jul 17 at 6:44

Karthik Sankar

13

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 6:58

N. F. Taussig

38.3k93053

edited Jul 17 at 6:58

N. F. Taussig

38.3k93053

edited Jul 17 at 6:58

N. F. Taussig

38.3k93053

38.3k93053

asked Jul 17 at 6:44

Karthik Sankar

13

asked Jul 17 at 6:44

Karthik Sankar

13

asked Jul 17 at 6:44

Karthik Sankar

13

13

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The combinations with $5$ correspond to two-number combinations from $1,2,3,4,6$.
  – Lord Shark the Unknown
  Jul 17 at 6:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So what you're asking is: When choosing $k$ items from a set of length $n$, how many of them include the item number $l$, so that the order does not matter? You can think about it like this: since the order does not matter, you can first choose the element that you are listing the occurrance of. After that, you are choosing $k-1$ items from a list of $n-1$.
  – Matti P.
  Jul 17 at 6:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @LordSharktheUnknown You mean from the set $1,2,3,4$, and indeed $4 choose 2 = 6$
  – Brevan Ellefsen
  Jul 17 at 6:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MattiP. Sorry for the inconvenience. I can't understand what you are pointing to n. I've updated the question. Can u pls read again!
  – Karthik Sankar
  Jul 17 at 6:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, I see that our notations are contradicting. What you call $n$ in the question is my $l$. What I mean by $n$ is the length of the list, which also has the fixed value $n=5$ in this case.
  – Matti P.
  Jul 17 at 7:38
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The combinations with $5$ correspond to two-number combinations from $1,2,3,4,6$.
  – Lord Shark the Unknown
  Jul 17 at 6:46
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  So what you're asking is: When choosing $k$ items from a set of length $n$, how many of them include the item number $l$, so that the order does not matter? You can think about it like this: since the order does not matter, you can first choose the element that you are listing the occurrance of. After that, you are choosing $k-1$ items from a list of $n-1$.
  – Matti P.
  Jul 17 at 6:48
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @LordSharktheUnknown You mean from the set $1,2,3,4$, and indeed $4 choose 2 = 6$
  – Brevan Ellefsen
  Jul 17 at 6:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MattiP. Sorry for the inconvenience. I can't understand what you are pointing to n. I've updated the question. Can u pls read again!
  – Karthik Sankar
  Jul 17 at 6:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, I see that our notations are contradicting. What you call $n$ in the question is my $l$. What I mean by $n$ is the length of the list, which also has the fixed value $n=5$ in this case.
  – Matti P.
  Jul 17 at 7:38
  

  
  
  
  

1

1

The combinations with $5$ correspond to two-number combinations from $1,2,3,4,6$.
– Lord Shark the Unknown
Jul 17 at 6:46

The combinations with $5$ correspond to two-number combinations from $1,2,3,4,6$.
– Lord Shark the Unknown
Jul 17 at 6:46

4

4

So what you're asking is: When choosing $k$ items from a set of length $n$, how many of them include the item number $l$, so that the order does not matter? You can think about it like this: since the order does not matter, you can first choose the element that you are listing the occurrance of. After that, you are choosing $k-1$ items from a list of $n-1$.
– Matti P.
Jul 17 at 6:48

So what you're asking is: When choosing $k$ items from a set of length $n$, how many of them include the item number $l$, so that the order does not matter? You can think about it like this: since the order does not matter, you can first choose the element that you are listing the occurrance of. After that, you are choosing $k-1$ items from a list of $n-1$.
– Matti P.
Jul 17 at 6:48

1

1

@LordSharktheUnknown You mean from the set $1,2,3,4$, and indeed $4 choose 2 = 6$
– Brevan Ellefsen
Jul 17 at 6:57

@LordSharktheUnknown You mean from the set $1,2,3,4$, and indeed $4 choose 2 = 6$
– Brevan Ellefsen
Jul 17 at 6:57

@MattiP. Sorry for the inconvenience. I can't understand what you are pointing to n. I've updated the question. Can u pls read again!
– Karthik Sankar
Jul 17 at 6:57

@MattiP. Sorry for the inconvenience. I can't understand what you are pointing to n. I've updated the question. Can u pls read again!
– Karthik Sankar
Jul 17 at 6:57

Ah, I see that our notations are contradicting. What you call $n$ in the question is my $l$. What I mean by $n$ is the length of the list, which also has the fixed value $n=5$ in this case.
– Matti P.
Jul 17 at 7:38

Ah, I see that our notations are contradicting. What you call $n$ in the question is my $l$. What I mean by $n$ is the length of the list, which also has the fixed value $n=5$ in this case.
– Matti P.
Jul 17 at 7:38

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote

If you are choosing combinations of k numbers from n,

each chosen number will have the same probability of occurrence, $frackn$

Thus the number of occurrences of any of them will be $binomnkcdotfrackn$

For your example, $binom53cdotfrac35 =6$

share|cite|improve this answer

answered Jul 17 at 6:57

true blue anil

20.3k11841

add a comment | 

up vote
2
down vote

You are taking $3$ numbers out of $5$ numbers. According to you, the place of $5$ must be fixed. So, there are two places remaining with $4$ possible numbers .

(I) In how many ways can you fill up the two places?

share|cite|improve this answer

edited Jul 17 at 7:02

answered Jul 17 at 6:56

Entrepreneur

16019

add a comment | 

up vote
1
down vote

If the list has $n$ numbers then there are $binomnk$ combinations of $k$ numbers.

So in total $binomnkk$ numbers occur in the combinations.

Then by symmetry each of $n$ numbers will occur: $$frac1nbinomnkk=binomn-1k-1$$times.

share|cite|improve this answer

answered Jul 17 at 8:16

drhab

86.6k541118

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

If you are choosing combinations of k numbers from n,

each chosen number will have the same probability of occurrence, $frackn$

Thus the number of occurrences of any of them will be $binomnkcdotfrackn$

For your example, $binom53cdotfrac35 =6$

share|cite|improve this answer

answered Jul 17 at 6:57

true blue anil

20.3k11841

add a comment | 

up vote
2
down vote

If you are choosing combinations of k numbers from n,

each chosen number will have the same probability of occurrence, $frackn$

Thus the number of occurrences of any of them will be $binomnkcdotfrackn$

For your example, $binom53cdotfrac35 =6$

share|cite|improve this answer

answered Jul 17 at 6:57

true blue anil

20.3k11841

add a comment | 

up vote
2
down vote

up vote
2
down vote

If you are choosing combinations of k numbers from n,

each chosen number will have the same probability of occurrence, $frackn$

Thus the number of occurrences of any of them will be $binomnkcdotfrackn$

For your example, $binom53cdotfrac35 =6$

share|cite|improve this answer

answered Jul 17 at 6:57

true blue anil

20.3k11841

If you are choosing combinations of k numbers from n,

each chosen number will have the same probability of occurrence, $frackn$

Thus the number of occurrences of any of them will be $binomnkcdotfrackn$

For your example, $binom53cdotfrac35 =6$

share|cite|improve this answer

answered Jul 17 at 6:57

true blue anil

20.3k11841

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 6:57

true blue anil

20.3k11841

answered Jul 17 at 6:57

true blue anil

20.3k11841

answered Jul 17 at 6:57

true blue anil

20.3k11841

20.3k11841

add a comment | 

add a comment | 

up vote
2
down vote

You are taking $3$ numbers out of $5$ numbers. According to you, the place of $5$ must be fixed. So, there are two places remaining with $4$ possible numbers .

(I) In how many ways can you fill up the two places?

share|cite|improve this answer

edited Jul 17 at 7:02

answered Jul 17 at 6:56

Entrepreneur

16019

add a comment | 

up vote
2
down vote

You are taking $3$ numbers out of $5$ numbers. According to you, the place of $5$ must be fixed. So, there are two places remaining with $4$ possible numbers .

(I) In how many ways can you fill up the two places?

share|cite|improve this answer

edited Jul 17 at 7:02

answered Jul 17 at 6:56

Entrepreneur

16019

add a comment | 

up vote
2
down vote

up vote
2
down vote

You are taking $3$ numbers out of $5$ numbers. According to you, the place of $5$ must be fixed. So, there are two places remaining with $4$ possible numbers .

(I) In how many ways can you fill up the two places?

share|cite|improve this answer

edited Jul 17 at 7:02

answered Jul 17 at 6:56

Entrepreneur

16019

You are taking $3$ numbers out of $5$ numbers. According to you, the place of $5$ must be fixed. So, there are two places remaining with $4$ possible numbers .

(I) In how many ways can you fill up the two places?

share|cite|improve this answer

edited Jul 17 at 7:02

answered Jul 17 at 6:56

Entrepreneur

16019

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 7:02

edited Jul 17 at 7:02

edited Jul 17 at 7:02

answered Jul 17 at 6:56

Entrepreneur

16019

answered Jul 17 at 6:56

Entrepreneur

16019

answered Jul 17 at 6:56

Entrepreneur

16019

16019

add a comment | 

add a comment | 

up vote
1
down vote

If the list has $n$ numbers then there are $binomnk$ combinations of $k$ numbers.

So in total $binomnkk$ numbers occur in the combinations.

Then by symmetry each of $n$ numbers will occur: $$frac1nbinomnkk=binomn-1k-1$$times.

share|cite|improve this answer

answered Jul 17 at 8:16

drhab

86.6k541118

add a comment | 

up vote
1
down vote

If the list has $n$ numbers then there are $binomnk$ combinations of $k$ numbers.

So in total $binomnkk$ numbers occur in the combinations.

Then by symmetry each of $n$ numbers will occur: $$frac1nbinomnkk=binomn-1k-1$$times.

share|cite|improve this answer

answered Jul 17 at 8:16

drhab

86.6k541118

add a comment | 

up vote
1
down vote

up vote
1
down vote

If the list has $n$ numbers then there are $binomnk$ combinations of $k$ numbers.

So in total $binomnkk$ numbers occur in the combinations.

Then by symmetry each of $n$ numbers will occur: $$frac1nbinomnkk=binomn-1k-1$$times.

share|cite|improve this answer

answered Jul 17 at 8:16

drhab

86.6k541118

If the list has $n$ numbers then there are $binomnk$ combinations of $k$ numbers.

So in total $binomnkk$ numbers occur in the combinations.

Then by symmetry each of $n$ numbers will occur: $$frac1nbinomnkk=binomn-1k-1$$times.

share|cite|improve this answer

answered Jul 17 at 8:16

drhab

86.6k541118

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 8:16

drhab

86.6k541118

answered Jul 17 at 8:16

drhab

86.6k541118

answered Jul 17 at 8:16

drhab

86.6k541118

86.6k541118

add a comment | 

add a comment | 

 

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