Number of partitions of a number I've been trying to find out how many partitions of a number I can make if I restrict the partition size, The specific problem I am trying to tackle is,
How many ways can you partition the number '9' when each partition has a size less than or equal 6 but greater than 0 (note: all partitions have a discrete size ( can't have a decimal or fractional size)
My attempt:
let A+B+C=9 and let $(A,B,C)\subset S^3$ where $S= (x: where \ 6\ge x\ge 0)$
and then I did the stars and bars approach to finding the number of partitions $\binom{11}{2} $ but I am stuck on how to remove the partitions of where one of the partitions has is greater than 6. Is there any way to directly arrive at the answer required or more preferably a way to fix this formula?
 A: I read your question as: "On how many ways can we write $9$ as a sum of elements of $\{1,2,3,4,5,6\}$? where the order of the terms does not matter."

If $T(n,k)$ denotes the number of ways that we can write non-negative integer $n$ as a sum of elements of $\{1,\dots,k\}$ then we have the recursion equality:$$T(n,k)=T(n,k-1)+T(n-k,k)$$The first term corresponds with the number of ways we can write $n$ as a sum of elements of $\{1,\dots,k-1\}$.
The second term corresponds with the number of ways we can write $n$ as a sum of elements of $\{1,\dots,k\}$ and moreover $k$ is one of the terms of this sum.
Also have a look Eulers triangular table at A026820-OEIS where I find that: $$T(9,6)=26$$
A: You want the number of partitions of $9$ into any parts, minus those with a part of size $7$, $8$ and $9$. So the number of partitions of $9$ is simply $30$. The number of partitions with a part of size at least $7$ is clearly
$$(9),\;\;(8,1),\;\;(7,2),\;\;(7,1,1)$$
so four. Thus there are $26$ partitions remaining.
