List all size $4$ subsets from the set $\{A,B,C,D,E,F\}$

Question

List all size $4$ subsets from the set $\{A,B,C,D,E,F\}$

So first I realized this is a permutation because subsets are the same off they have the same elements. So to figure out how many subsets there would be 6p4 which is 15. However I am having trouble listing the actual subsets. Is there a more organized way to go about this so I can ensure I’m not missing any? This is what I have so far:

$\{A,B,C,D\} \{B,C,D,E\} \{C,D,E,F\} \{D,E,F,A\} \{E,F,A,B\} \{F,A,B,C\} \{B,C,D,F\} \{A,B,E,F\} \{A,B,D,F\} \{C,D,A,F\}$

Answer 1

This might make your life slightly easier a bit.

Choosing $4$ elements out of $6$ is equivalent to choose $2$ elements out of $6$ and then discard them.

For example if I were to discard $\{A, B\}$, the set that I am listing is $\{A,B\}^c=\{C,D,E,F\}$.

Do the same thing as the rest $AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF$.

In some sense, you just have to write out $A$ to $F$ and systematically cover $2$ letters at the same time using two fingers to list out all the subset of cardinality $4$.

Answer 2

I would use alphabetical order within each subset (ADEF, rather than DEFA), and overall I'd put the subset strings in alphabetical order:

ABCD

ABCE

ABCF

ABDE

etc.

Answer 3

If you only have pen and paper you can do this: let's write all subsets in a table $\binom{n}{k} \times k$, such that each row describes a subset and elements of a row are ordered in ascending order. Let's fill in the table column-by-column. The first column is easy: there are $\binom{n-1}{k-1}$ subsets with the smallest element 'A', $\binom{n-2}{k-1}$ subsets with the smallest element 'B', $\binom{n-3}{k-1}$ subsets with the smallest element $C$ and so on. In that way, your table is separated into $n-k+1$ subtables each of them having the first column containing equal elements. So now you can perform the same procedure for each of them. Let's do it for $n=6;\;k=4$: $\binom{5}{3} = 10$, $\binom{4}{3} = 4$, $\binom{3}{3} = 1$, so the first column is

AAAAAAAAAABBBBC

For the first subtable (with A) $\binom{4}{2} = 6$, $\binom{3}{2} = 3$, $\binom{2}{2} = 1$ so the second column looks like

AAAAAAAAAABBBBC
BBBBBBCCCD_____

Let's fill the blanks, for the second subtable $\binom{3}{2} = 3$, $\binom{2}{2} = 1$; for the third subtable $\binom{2}{2} = 1$ which gives us:

AAAAAAAAAABBBBC
BBBBBBCCCDCCCDD

If we continue like this we will get:

AAAAAAAAAABBBBC
BBBBBBCCCDCCCDD
CCCDDEDDEEDDEEE
DEFEFFEFFFEFFFF

Answer 4

We need to use "combinations", not permutations.

So we need to use "6 choose 4", or $\dbinom 64 = \dfrac{6!}{4!2!} = \dfrac{6\cdot 5}{2} = 15$ such subsets (combinations) of $\{A, B, C, D, E, F\}$

Start with all four element subsets starting with $A$:

ABC... (3) $\quad \{A, B, C, D\},\;\;\{A, B, C, E\}, \;\;\{A, B, C, F\}$

ABD.. (2) $\quad \{A, B, D, E\}, \;\;\{A, B, D, F\}$

ABEF

ACD... (2) $\quad \{A, C, D, E\},\;\; \{A, C, D, F\}$

ACEF

ADEF

BCD... (2) $\quad \{B, C, D, E\}, \;\; \{B, C, D, F\}$

BCEF

BDEF,

CDEF

$$\{A, B, C, D\}, \{A, B, C, E\}, \{A, B, C, F\}, \{A, B, D , E\},$$ $$ \{ A, B, D, F\}, \{A, B, E, F\}, \{A, C, D, E\}, \{A, C, D, F\}, $$ $$\{A, C, E, F\}, \{A, D, E, F\}, \{B, C, D, E\}, \{B, C, D, F\},$$ $$ \{B, C, E, F\}, \{B, D, E, F\}, \{C, D, E, F\}$$