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Algebra II HW Help

S

Shadow Conception

Golden Member
"If a set contains N elements, what is the total number of subsets that the set has?"

Can somebody explain this to me?

I know that if a set contains elements {a, b, c}, then you end up with subsets {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} and {}. But how can I relate this to a variable?
 
I don't recall ever seeing this in any of my math classes... and this includes Cal 1-3 and DiffEQ, much less Algebra II
 
Originally posted by: fatpat268
I don't recall ever seeing this in any of my math classes... and this includes Cal 1-3 and DiffEQ, much less Algebra II
Click to expand...
Eh, really? I remember learning permutations in my 9th grade Algebra II class.
 
2^n.

Say N = 1; you have {a}. {}. So you have 2 = 2^1 = 2
Say N = 2; you have {a}, {b}, {ab}. {}. So you have 4 = 2^2 = 4
Say N = 3; you have {a}, {b}, {c}, {ab}, {ac}. {bc}, {abc}. {}. So you have 8 = 2^3 = 8
Say N = 4; you have {a}, {b}, {c}, {d}, {ab}, {ac}, {ad}, {bc}, {bd}, {cd}, {abc}, {abd}, {acd}, {bcd}, {abcd}, {}. So you have 16 = 2^4 = 16.

If you want the exact reason why, scroll to the bottom of the page.

http://www.tutorvista.com/cont...ts/results-subsets.php

The easiest way to understand it is the following explanation I ripped from a page:

For a set of n there are 2^n subsets. This is easily proved: Any subset of the set can either contain or not contain an element; so, for a subset, there are 2 states for the first element, 2 for the second element, ?, 2 for the nth element; so, there are 2 states for the first element, 2 * 2 = 2^2 states for the first two, 2 * 2 * 2= 2^3 states for the first three, ?, 2 * 2 * 2 * ? * 2 = 2^n states for all the n elements.
Click to expand...
 
Originally posted by: BigJ
2^n.

Say N = 1; you have {a}. {}. So you have 2 = 2^1 = 2
Say N = 2; you have {a}, {b}, {ab}. {}. So you have 4 = 2^2 = 4
Say N = 3; you have {a}, {b}, {c}, {ab}, {ac}. {bc}, {abc}. {}. So you have 8 = 2^3 = 8
Say N = 4; you have {a}, {b}, {c}, {d}, {ab}, {ac}, {ad}, {bc}, {bd}, {cd}, {abc}, {abd}, {acd}, {bcd}, {abcd}, {}. So you have 16 = 2^4 = 16.

If you want the exact reason why, scroll to the bottom of the page.

http://www.tutorvista.com/cont...ts/results-subsets.php

The easiest way to understand it is the following explanation I ripped from a page:

For a set of n there are 2^n subsets. This is easily proved: Any subset of the set can either contain or not contain an element; so, for a subset, there are 2 states for the first element, 2 for the second element, ?, 2 for the nth element; so, there are 2 states for the first element, 2 * 2 = 2^2 states for the first two, 2 * 2 * 2= 2^3 states for the first three, ?, 2 * 2 * 2 * ? * 2 = 2^n states for all the n elements.
Click to expand...
Click to expand...

Or you could use induction (this is "thought induction", not strictly mathematical induction):

Say you have a set of n elements, and say that it has m possible subsets. Then, adding an element, e to the set, you will have twice as many possible subsets, since you can break down the possible subsets into subsets with e and subsets without e. Subsets without e are the same as the subsets when you only had n elements, and subsets with e are the same, except they each have element e in them. So, you have 2*m subsets. If you work backwards, starting at the empty set (1 possible subset), you see the size double each time, and thus follow the 2^n formula.
 
Originally posted by: BigJ
The easiest way to understand it is the following explanation I ripped from a page:

For a set of n there are 2^n subsets. This is easily proved: Any subset of the set can either contain or not contain an element; so, for a subset, there are 2 states for the first element, 2 for the second element, ?, 2 for the nth element; so, there are 2 states for the first element, 2 * 2 = 2^2 states for the first two, 2 * 2 * 2= 2^3 states for the first three, ?, 2 * 2 * 2 * ? * 2 = 2^n states for all the n elements.
Click to expand...
Click to expand...

This is the way to go.

Think about a problem from elementary school. You have 3 pairs of pants and 4 shirts, how many outfits do you have? Clearly the answer is (3 options) x (4 options) = 12. The subset problem is the same.

You have n elements {E1, E2, E3, ... , En}

For each element you have two choices, include or not include. Similar to the pants and shirts problems you simply multiply your options together.

(2 options for E1) x (2 options for E2) x ... x (2 options for En) = 2^n
 
Originally posted by: Shadow Conception
"If a set contains N elements, what is the total number of subsets that the set has?"

Can somebody explain this to me?

I know that if a set contains elements {a, b, c}, then you end up with subsets {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} and {}. But how can I relate this to a variable?
Click to expand...

2^N

Each element can be in a subset, or not in a subset ... therefore two possibilities for each of the N elements, therefore 2^N.
 
Originally posted by: blinky8225
Originally posted by: fatpat268
I don't recall ever seeing this in any of my math classes... and this includes Cal 1-3 and DiffEQ, much less Algebra II
Click to expand...
Eh, really? I remember learning permutations in my 9th grade Algebra II class.
Click to expand...

Nope, I didn't even know what permutations were (well, I do now, I just read up on it)

Originally posted by: TuxDave
Originally posted by: fatpat268
I don't recall ever seeing this in any of my math classes... and this includes Cal 1-3 and DiffEQ, much less Algebra II
Click to expand...

Pay attention in class!
Click to expand...

I did pay attention. In fact, I'm a math tutor at my college, and there aren't permutations even in this college algebra book they use. Or if there was, I've been over looking it the entire time.
 
Originally posted by: fatpat268
Originally posted by: blinky8225
Originally posted by: fatpat268
I don't recall ever seeing this in any of my math classes... and this includes Cal 1-3 and DiffEQ, much less Algebra II
Click to expand...
Eh, really? I remember learning permutations in my 9th grade Algebra II class.
Click to expand...

Nope, I didn't even know what permutations were (well, I do now, I just read up on it)

Originally posted by: TuxDave
Originally posted by: fatpat268
I don't recall ever seeing this in any of my math classes... and this includes Cal 1-3 and DiffEQ, much less Algebra II
Click to expand...

Pay attention in class!
Click to expand...

I did pay attention. In fact, I'm a math tutor at my college, and there isn't permutations even in this college algebra book they use. Or if there was, I've been over looking it the entire time.
Click to expand...

Permutations probably weren't used much in any of the classes you used. It was probably mentioned briefly in algebra and/or calculus.

I didn't use set theory enough to remember this stuff until I took discrete math
 
Originally posted by: Leros
Permutations probably weren't used much in any of the classes you used. It was probably mentioned briefly in algebra and/or calculus.

I didn't use set theory enough to remember this stuff until I took discrete math
Click to expand...

Same here.
 
Originally posted by: chuckywang
Originally posted by: Shadow Conception
"If a set contains N elements, what is the total number of subsets that the set has?"

Can somebody explain this to me?

I know that if a set contains elements {a, b, c}, then you end up with subsets {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} and {}. But how can I relate this to a variable?
Click to expand...

2^N

Each element can be in a subset, or not in a subset ... therefore two possibilities for each of the N elements, therefore 2^N.
Click to expand...

This

God, I hate "manglish," my coined phrase for the terms used for definitions and proofs. They have the same relationship to math that grammar has to English i.e. none to speak of.
 
Yikes! Permutations are in 9th grade math in NY.
(not that this is a permutation)

 
I am going to threadjack here.

If you have 10 pills, 6 reds and 4 blues, what is the probability you will get all 4 blue pills when you take 6 pills?
 
The numeral 2 is unpleasant; it looks like a twisted coat hanger. We need a better glyph to represent two.
 
Originally posted by: Shadow Conception
"If a set contains N elements, what is the total number of subsets that the set has?"

Can somebody explain this to me?

I know that if a set contains elements {a, b, c}, then you end up with subsets {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} and {}. But how can I relate this to a variable?
Click to expand...

It's true what they say.... you're never going to have to know this stuff post-college.
 
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