# Find number of occurrences of $n$ in a combination

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?

• The combinations with $5$ correspond to two-number combinations from $\{1,2,3,4,6\}$. – Lord Shark the Unknown Jul 17 '18 at 6:46
• 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 '18 at 6:48
• @LordSharktheUnknown You mean from the set $\{1,2,3,4\}$, and indeed ${4 \choose 2} = 6$ – Brevan Ellefsen Jul 17 '18 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 '18 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 '18 at 7:38

If you are choosing combinations of k numbers from n,
each chosen number will have the same probability of occurrence, $\frac{k}{n}$

Thus the number of occurrences of any of them will be $\binom{n}{k}\cdot\frac{k}{n}$

For your example, $\binom53\cdot\frac35 =6$

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?

If the list has $n$ numbers then there are $\binom{n}{k}$ combinations of $k$ numbers.

So in total $\binom{n}{k}k$ numbers occur in the combinations.

Then by symmetry each of $n$ numbers will occur: $$\frac1n\binom{n}{k}k=\binom{n-1}{k-1}$$times.