Which way is true?

$$A = \{1,2,3,4,5,6\}$$

How many different subsets that don't involve $4$ and $5$ can be written?

I've solved this question with two part.

Assumed that

$$A = \{1,2,3,6\}$$

By the way

$$2^4 = 16 \tag {1}$$

The other way I've used

$$A = \{1,2,3,4,5,6\} = 2^6$$ and

$$A = \{1,2,3,6\} = 2^4$$

Then

$$2^6 - 2^4$$

$$64 - 16 = 48 \tag {2}$$

So, I've got two answer, $48$ and $16$. Which way is true then?

• What would lead you to compute $2^6-2^4$? (your first method looks fine)
– lulu
Nov 6, 2017 at 19:22
• According to my teacher, the first method isn't correct. He accepted the second method. Nov 6, 2017 at 19:23
• If you wanted to start from the $2^6$ subsets of the big set you would need to subtract all those that contain a $5$ then all those that contain a $6$ then add back all those that contain both. Good exercise to verify that you get $2^4$ again.
– lulu
Nov 6, 2017 at 19:24
• Are the both true? Why did he accept the second method? Nov 6, 2017 at 19:24
• List: {}, {1}, {2}, {3}, {6}, {1, 2}, {1, 3}, {1, 6}, {2, 3}, {2, 6}, {3, 6}, {1, 2, 3}, {1, 2, 6}, {1, 3, 6}, {2, 3, 6}, {1, 2, 3, 6} Nov 6, 2017 at 19:28

Subsets of $A=\{1,2,3,4,5,6\}$

$$\mathscr{P}(A)=\{\{\},\{1\},\{2\},\{3\},\{4\},\{5\},\{6\},\{1,2\},\{1,3\},\{1,4\},\{1,5\},\{1,6\},\{2,3\},\{2,4\},\{2,5\},\{2,6\},\{3,4\},\{3,5\},\{3,6\},\{4,5\},\{4,6\},\{5,6\},\{1,2,3\},\{1,2,4\},\{1,2,5\},\{1,2,6\},\{1,3,4\},\{1,3,5\},\{1,3,6\},\{1,4,5\},\{1,4,6\},\{1,5,6\},\{2,3,4\},\{2,3,5\},\{2,3,6\},\{2,4,5\},\{2,4,6\},\{2,5,6\},\{3,4,5\},\{3,4,6\},\{3,5,6\},\{4,5,6\},\{1,2,3,4\},\{1,2,3,5\},\{1,2,3,6\},\{1,2,4,5\},\{1,2,4,6\},\{1,2,5,6\},\{1,3,4,5\},\{1,3,4,6\},\{1,3,5,6\},\{1,4,5,6\},\{2,3,4,5\},\{2,3,4,6\},\{2,3,5,6\},\{2,4,5,6\},\{3,4,5,6\},\{1,2,3,4,5\},\{1,2,3,4,6\},\{1,2,3,5,6\},\{1,2,4,5,6\},\{1,3,4,5,6\},\{2,3,4,5,6\},\{1,2,3,4,5,6\}\}$$

Subsets of $B=\{1,2,3,6\}$

$$\mathscr{P}(B)=\{\{\},\{1\},\{2\},\{3\},\{6\},\{1,2\},\{1,3\},\{1,6\},\{2,3\},\{2,6\},\{3,6\},\{1,2,3\},\{1,2,6\},\{1,3,6\},\{2,3,6\},\{1,2,3,6\}\}$$

$$\mathscr{P}(A)\setminus \mathscr{P}(B)=\{\{4\},\{5\},\{1,4\},\{1,5\},\{2,4\},\{2,5\},\{3,4\},\{3,5\},\{4,5\},\{4,6\},\{5,6\},\{1,2,4\},\{1,2,5\},\{1,3,4\},\{1,3,5\},\{1,4,5\},\{1,4,6\},\{1,5,6\},\{2,3,4\},\{2,3,5\},\{2,4,5\},\{2,4,6\},\{2,5,6\},\{3,4,5\},\{3,4,6\},\{3,5,6\},\{4,5,6\},\{1,2,3,4\},\{1,2,3,5\},\{1,2,4,5\},\{1,2,4,6\},\{1,2,5,6\},\{1,3,4,5\},\{1,3,4,6\},\{1,3,5,6\},\{1,4,5,6\},\{2,3,4,5\},\{2,3,4,6\},\{2,3,5,6\},\{2,4,5,6\},\{3,4,5,6\},\{1,2,3,4,5\},\{1,2,3,4,6\},\{1,2,3,5,6\},\{1,2,4,5,6\},\{1,3,4,5,6\},\{2,3,4,5,6\},\{1,2,3,4,5,6\}\}$$

If we are looking for the subsets that do not contain $4$ or $5$ or both then the answer is $16$

Hope this helps

• The main issue with the question was the wording - does it mean "doesn't contain 4 or 5" or does it mean "doesn't contain both 4 and 5? Nov 6, 2017 at 22:41
• Why did he accept the second way? Nov 7, 2017 at 13:40
• @Maxime Ask him. It is a linguistic issue. Furthermore my English is pretty poor... Nov 7, 2017 at 15:01