Possible Partitions

By the given additional property we can say that {1,2,4,8} , {3,6} and {5,10} will be in same set. Therefore Let X= {1,2,4,8} , Y = {3,6}, Z={5,10} , L ={ 7}, K= {9}

So number of combinations = No of ways of distributing X, Y, Z,L,K into 3 non-empty sets.

This can be identified as stirling number S(5,3)= 25

Since the additional property is x and 2x must be in the same group if 2x \leq 10, there are 5 subsets of numbers from 1 through 10 that always in the same group, they are: {1,2,4,8}, {3,6}, {5,10}, {7}, {9}. Hence, we only need to count the combination of those 5 subsets. Since the subsets are divided into three groups, we could merely divide them into two possibilities, they are 3-1-1 and 2-2-1. - 3-1-1 To find the number of this possibility, we use the concept of combination. It is similar to make 3 of 5 subsets be in the same group, while the other 2 subsets are made separated (1-1). But, remember that we have to divide the result by 2!, because there is two groups with the same number of subsets. Hence, we have: \frac {{5 \choose 3} \times {2 \choose 1}}{2!} = 10 - 2-2-1 We make 2 of 5 subsets be in the same group, 2 of 3 remaining subsets also in the same group, and another subset in the other group. But, remember that we have to divide the result by 2!, because there is two groups with the same number of subsets. Hence, we have: \frac{{5 \choose 2} \times {3 \choose 2}}{2!} = 15

Therefore, the number of different ways to create yhe groups are 10 + 15 = 25

Given the "additional property", we can form sets of numbers which must be in the same group: $\{1,2,4,8\},\{3,6\},\{5,10\},\{7\},\{9\}$ . So these $5$ sets can be distributed among the $3$ groups, with each group getting at least $1$ set. It is easy to see that we either have $2$ groups with $1$ set and $1$ group with $3$ sets, or have $2$ groups with $2$ sets and $1$ group with $1$ set. Since the groups are not labelled, the former case gives ${5 \choose 3} = 10$ possibilities and the latter case gives $\dfrac{{5 \choose 2}\cdot {3\choose 2}}{2} = 15$ possibilities. There are thus $10+15=25$ different ways to create the groups.

Any integer $n$ can be expressed as $n = 2^{k}m$ where $m$ is odd. If $n \leq 10$ then $m \leq 10$ as well, and so $m$ and $n$ must be in the same group. Thus, we only need consider what groups the odd numbers are in, and the groups for the even numbers will be determined. There are five odd numbers between 1 and 10, so we have 5 numbers to distribute among three groups. We can either have one group of three and two groups of one, or one group of one and two groups of two. There are $\binom{5}{2} = 10$ ways to put them in two groups with one and one group with three. There are $\binom{5}{2}\times \binom{3}{2} / 2! = 15$ ways to arrange them in two groups of two and one group of one. Thus there are a total of $10+15 = 25$ ways to arrange them.