Fundamental Principle of Counting

Let

$$x_1=\text{number of elements belonging to exactly one of }A,B,C$$

$$x_2=\text{number of elements belonging to exactly two of }A,B,C$$

Since $$A\cap B\cap C=\emptyset$$

there are no elements belonging to all three sets.

Also, $$A\cup B\cup C=X$$ and $$|X|=7$$

Hence $$x_1+x_2=7$$

Now count total memberships.

Since $$|A|=|B|=|C|=3$$ the total number of memberships is $$|A|+|B|+|C|=9$$

On the other hand, each element counted in exactly one set contributes 1,and each element counted in exactly two sets contributes 2.

Therefore,

$$x_1+2x_2=9$$

Subtracting

$$x_1+x_2=7$$ from $$x_1+2x_2=9$$ gives $$x_2=2$$

Hence

$$x_1=5$$

So among the 7 elements:

- 5 elements belong to exactly one set,

- 2 elements belong to exactly two sets.

First choose the 2 elements that belong to exactly two sets:

$$\binom72=21$$

For each of these 2 elements, choose which pair of sets contains it.

Possible pairs are

$$(A,B),\ (B,C),\ (C,A)$$

Thus

$$3^2$$

choices.

Now consider the remaining 5 elements.

Each must belong to exactly one of

$$A,\ B,\ C$$

Let

$$n_A,n_B,n_C$$

be the numbers assigned exclusively to

$$A,B,C$$

respectively.

The two doubly-assigned elements contribute a total of 4 memberships.

Since total memberships in each set must be 3, let

$$d_A,d_B,d_C$$

be the number of times the doubly-assigned elements occur in

$$A,B,C$$

respectively.

Then

$$n_A=3-d_A$$

$$n_B=3-d_B$$

$$n_C=3-d_C$$

and

$$n_A+n_B+n_C=5$$

because there are 5 singly-assigned elements.

The pair-types of the two doubly-assigned elements determine

$$d_A,d_B,d_C$$.

Count according to the distribution of the two pair-types.

Case 1: Both doubly-assigned elements use the same pair.

There are

$$3$$

choices.

For example, both are of type

$$(A,B)$$.

Then

$$d_A=d_B=2,\quad d_C=0$$

Hence

$$n_A=1,\quad n_B=1,\quad n_C=3$$

Number of assignments of the 5 single elements:

$$\frac{5!}{1!1!3!}=20$$

Contribution:

$$3\times20=60$$

Case 2: The two doubly-assigned elements use different pairs.

Choose ordered pair-types:

$$3\times2=6$$

choices.

For example,

$$(A,B)$$

and

$$(A,C)$$.

Then

$$d_A=2,\quad d_B=1,\quad d_C=1$$

Hence

$$n_A=1,\quad n_B=2,\quad n_C=2$$

Number of assignments:

$$\frac{5!}{1!2!2!}=30$$

Contribution:

$$6\times30=180$$

Total assignments for the remaining 5 elements:

$$60+180=240$$

Finally,

$$\text{Total}=\binom72\times240$$

$$=21\times240$$

$$=5040$$

Final Answer :

$$5040$$