Help me find the cardinality of this set anons

>>8481140
It can't be empty because 7 is in it.

>CS majors

>>8481156
can you elaborate more please?
how can 7 be a subset of AΛ{3,5}∩A=O

File: 1427205095887.png (2MB, 1372x909px) Image search: [Google]

2MB, 1372x909px

alright I think I might have a solution.

I'll motivate my solution by considering the cases.

the first case is sets with 1 element, this set contains only seven, hence there is only one such set.

the second case is the set containing 2 elements, this set contains seven and some other element which is not seven, three, nor five, there being (n-3) such choices, hence there are (n-3) sets of this size

the next case is are the sets containing three elements, which all have seven, and then a choice of (n-3) then a choice of (n-4)

and so on.

This leads to a general formula which looks something like this

[eqn]1+\sum_{j=3}^{n}\prod_{k=3}^{j}(n-k)=|X|[/eqn]

where X is the set in question, this may then be proven bu induction or something, but I think the argument I've provided is atleast heuristically compelling, if nothing else.


I could have made a mistake, I'm wrong all the time, so I make no guarantees.

>>8481200

oh yeah, and I think it should be stated that this is true where n>7, for n<7 the set is empty I think, sorry.

>>8481206
Thank you ill try find a solution based on what you said.

>>8481200

ahhh, the n on top of the sigma should probably be n-1

>>8481214

and the method does overcount, its wrong, unfortunately.

for n=7 you have
that every A is a subset of {1,2,4,6} U {7}
so you take the power set of {1,2,4,6} and you get 16 terms
for n+1 you get the double amount of terms and so on
i guess

>>8481140

ok, sorry, new guess

[eqn]\sum_{k=0}^{n-3} {n-3 \choose k}[/eqn]

is simpler and shouldn't over count

[eqn] 2^{n-3}[/eqn]
n>6

>>8481224
wow thanks ,it looks right gonna test it