Let a subset of integer S = { -4<a<15; a is integer}.We have make partition for S such that
- Partition will contains exactly 5 disjoint subsets.
2)Among these 5 subsets exactly one has 4 elements.
- Each subsets contains at least 2 elements. Question is : How many such partition possible ? My attempt so far : I consider the cardinality of the susets as X1,X2,...,X5 with X5=4 and try to solve the equation X1+X2+...+X5 = 18 ( Since 18 is the cardinality of S). This gives X1+...+X4=14 with Xi > 1 and Xi is not 4 for any i=1,2,3,4. Now i can apply the integral Solution of X1+...+Xk = n is (n+k-1) C (k-1) with the restriction property. But how do i compute that Xi is not 4. Please help.
That approach won’t work, because the elements of $S$ are distinguishable: different partitions have the same values of $x_1,\ldots,x_5$. I would start by noticing that there are $\binom{18}4$ ways to choose the $4$-element set. Once we’ve done that, we’re left with a set of $14$ elements, and we need to count the partitions of it into $4$ sets, each of which contains at least $2$ elements, and none of which contains $4$ elements.
Since $4\cdot3$ is only $12$, at least one part must have more than $4$ elements. The possible combinations of sizes for the sets in such a partition are:
$$\begin{align*} &8,2,2,2\\ &7,3,2,2\\ &6,4,2,2\\ &6,3,3,2\\ &5,5,2,2\\ &5,4,3,2\\ &5,3,3,3 \end{align*}$$
It’s tedious but not difficult to calculate the number of partitions there are of each type. For instance, there are
$$\frac{\binom{14}8\binom62\binom42\binom22}{3!}$$
partitions of the first type; the denominator of $3!$ reflects the fact that the same three parts of size $2$ can be chosen in $3!$ different orders to yield the same partition. Similarly, there are
$$\frac{\binom{14}7\binom73\binom42\binom22}{2!}$$
partitions of the second type. One can perform the same kind of calculation for each of the seven possibilities and take the sum to get the total number of acceptable partitions of the $14$ remaining elements of $S$. The final answer will be the product of $\binom{18}4$ and that sum, since each $4$-element subset of $S$ can be combined with any suitable partition of the rest of $S$.