# subset question

• January 23rd 2009, 10:32 AM
tukilala
subset question
A⊂(1,2,3......100) |A|=10
how many subsets posibilities exist that A will be a subset.
i know the answers is 2^10, but why????
if A⊂(1,2,3......200) |A|=10, so the number of posebilitis of A been a subset is still 2^10??? why? and how???
• January 23rd 2009, 11:31 AM
Plato
Quote:

Originally Posted by tukilala
A⊂(1,2,3......100) |A|=10
how many subsets posibilities exist that A will be a subset.
i know the answers is 2^10, but why???? if A⊂(1,2,3......200) |A|=10, so the number of posebilitis of A been a subset is still 2^10??? why? and how???

There are indeed $2^{10}$ subsets of A.
BUT, that is not the way you have worded the question.
You have ask “how many subsets of {1,2,3,…,100} are supersets of A?”
That is how many subsets of {1,2,3,…,100} have A as a subset.
The answer to that is $2^{90}$. So which is your question?
Is it about subsets of A? Or is it about supersets of A?
There is no point in my guessing which question you need help with.
• January 23rd 2009, 12:05 PM
tukilala
i ment how many sets with length=10 exist in {1,2,3,....,100}

for example: (1,2,3,4,5,6,7,8,9,10),(1,2,3,4,5,6,7,8,9,11),(1,1 0,20,30,40,50,60,70,80,90,100),(91,92,93,94,95,96, 97,98,99,100) ,.................................
so how many? and how are you get to the answer???
thanks
• January 23rd 2009, 12:08 PM
Plato
Quote:

Originally Posted by tukilala
i ment how many sets with length=10 exist in {1,2,3,....,100}

That is a simple combination question: ${{100}\choose{10}}=\frac{100!}{(10!)(90!)}$
• January 23rd 2009, 05:35 PM
tukilala
ok,
so i want to pick 10 elements of the initial set, which contains 100 elements
i know that the answer sepose to be 2^10, but why???? i dont understand...
if i want to pick 10 elements of the initial set, which contains 200 elements, so how many sub set will be now? still 2^10?? if yes, why?
if not so how many? and why?
thnx
• January 24th 2009, 04:24 AM
Plato
Quote:

Originally Posted by tukilala
ok,
so i want to pick 10 elements of the initial set, which contains 100 elements i know that the answer sepose to be 2^10, but why????

That statement is simply wrong!
There are $\frac{100!}{(90!)(10!)}$ ways to pick 10 elements from a set of 100.

Why do you think there are only $2^{10}$? That is the number of subsets in a set of 10.

There are $\frac{200!}{(190!)(10!)}$ ways to pick 10 elements from a set of 200.

There are $\frac{300!}{(290!)(10!)}$ ways to pick 10 elements from a set of 300.

etc