# ordered pairs $(A,B)$ of subsets of $X$ >such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is

Let $$X$$ be a set of $$5$$ elements. Then the number of ordered pairs $$(A,B)$$ of subsets of $$X$$

such that $$A\neq \phi\;,B\neq \phi$$ and $$A\cap B = \phi\;,$$ is

$$\bf{My\; Try::}$$ Let $$X = \left\{1,2,3,4,5\right\}\;,$$ then total no. of Subest of $$X = 2^5 = 32$$

(Which also contain $$\phi$$)

$$\bullet\;$$ If $$A = \{1\}\;,$$ Then $$B=\left (\{2\}\;,\{3\}\;,\{4\}\;,\{5\}\;,\{2,3\}\;,\{2,4\}\;,\{2,5\}\;,\{3,4\}\;,\{3,5\}\;,\{4,5\}\;,\{2,3,4\}\;,\{2,3,5\}\;,\{3,4,5\}\;,\{2,4,5\}\;,\{2,3,4,5\}\right )$$

Similarly for $$A=\{2\}\;,A=\{3\}\;,A=\{4\}$$ and $$A=\{5\}\;,$$ we get $$15$$ ordered pair for each single

element ed set $$A$$

$$\bullet\;$$ If $$A = \{1,2\}\;,$$ Then $$B=\left(\{3\}\;,\{4\}\;,\{5\},\{3,4\}\;,\{3,5\}\;,\{4,5\}\;,\{3,4,5\}\right)$$

Similarly for $$A=\{1,3\}\;,A=\{1,4\}\;,A=\{1,5\}\;,A=\{2,3\}\;,A=\{2,4\}\;,A=\{2,5\}\;,A=\{3,4\}\;,A=\{3,5\}\;,A=\{4,5\}$$ we get $$7$$ ordered pair for each double elemented set $$A$$

$$\bullet\;$$ If $$A=\{1,2,3\}\;,$$ Then $$B=\left(\{4\}\;,\{5\}\;,\{4,5\}\right)$$

So for $$10$$ ordered pair of $$A\;,$$ we get $$3$$ ordered pair of $$B$$

$$\bullet \;$$ If $$A = \{1,2,3,4\}\;,$$ Then $$B = \{5\}$$

So for $$5$$ ordered pair of $$A\;,$$ we get $$1$$ ordered pair of $$B$$

So Total ordered pair of $$\left(A,B\right)$$ is $$= \left(5\times 15\right)+\left(10 \times 7\right)+\left(10 \times 3\right)+\left(5 \times 1\right) = 75+70+30+5 = 180$$

If Question is How can we solve using Combination (selection) way.

plz explain me, Thanks

Phicar’s answer gives you a nice, short calculation if you know about Stirling numbers of the second kind. If not, you can still organize your argument a bit more efficiently.

Suppose that the set $A\cup B$ has $n$ elements; clearly $n$ must be $2,3,4$, or $5$. For each of these four possible values of $n$ we can argue as follows.

There are $\binom5n$ ways to choose the set $A\cup B$. $A$ can be any non-empty proper subset of $A\cup B$. $A\cup B$ has $2^n$ subsets, but one is empty and one is all of $A\cup B$, so there are only $2^n-2$ choices available for $A$. Thus, there are $\binom5n(2^n-2)$ ordered pairs $\langle A,B\rangle$ with $|A\cup B|=n$.

$$\sum_{n=2}^5\binom5n(2^n-2)=\sum_{n=2}^5\binom5n2^n-2\sum_{n=2}^5\binom5n\;.$$

Now notice that

$$\sum_{n=2}^5\binom5n=\sum_{n=0}^5\binom5n-\binom51-\binom50=2^5-5-1=26\;,$$

and

\begin{align*} \sum_{n=2}^5\binom5n2^n&=\sum_{n=0}^5\binom5n2^n-\binom512-\binom50\\\\ &=\sum_{n=0}^5\binom5n2^n1^{5-n}-10-1\\\\ &=(2+1)^5-11\\\\ &=232\;, \end{align*}

so the final answer is $232-52=180$.

This calculation suggests another elementary way to perform the calculation. If we temporarily allow $A$ and $B$ to be empty, we are in effect counting the ways to split $X$ into $3$ pieces, any of which may be empty. For each of the $5$ elements of $X$ we can put that element into $A$, into $B$, or into $X\setminus(A\cup B)$. This is a $3$-way choice made $5$ times, so there are $3^5=243$ ways to make it. However, some of these splits leave $A$ or $B$ or both empty. We can use an inclusion-exclusion argument to take care of them.

How many of these splits leave $A$ empty? They are the splits that put every element into $B$ or $X\setminus(A\cup B)$, so there are $2^5$ of them. There are also $2^5$ splits leaving $B$ empty, so we have to subtract $2\cdot2^5=64$. However, that subtracts the one split with $A=B=\varnothing$ twice, so we have to add it back in. The final result is

$$243-64+1=180\;.$$

Hi.
You might want to look about Stirling numbers of the second kind. You want the number of ways you can take a surjective function from $[5]=\{1,2,3,4,5\}$ to $[2]=\{1,2\}$ plus the number of surjective functions from $[5]=\{1,2,3,4,5\}$ to $[3]=\{1,2,3\}$.(in $3$ to represent elements that are not taken)
In your case $2!*S_{5,2}+3!*S_{5,3}=180$.
Hope this helps.

First let us try to get the number of ordered pairs $(A,B)$ such that $A\cup B = \phi$, without any restrictions on $A$ and $B$. Clearly, each element can EITHER be in only $A$ OR only $B$ OR neither. So, that is $3^5$. Now, remove the cases when $A = \phi$. This is $2^5-1$. Similarly for $B$. So, the total is $3^5 - 2^5 - 2^5 + 1 = 180$ (the +1 is because we have counted the pair $(A,B) = (\phi,\phi)$ twice).