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July 28, 2016
Posted by **Veronica** on Sunday, May 4, 2008 at 8:23am.

A question accured " M = {a,e,i,o,u}

a) how many subsets are there of M ?

When I did it was two , because a subset from my knowledge is that all of the elements in the set are there with the empty set

The answer was said at the back of the book , that it was wrong and the answers were "

{ a, e , o i u } ( a , o , e } {a , e ,u } ... etc , all of the answers that should be the " proper set " of M

can someone clarify wheither it's me who's wrong or the book

thanks

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**Damon**, Sunday, May 4, 2008 at 10:32amThere is 1 empty set

the combinations of 5 elements taken zero at a time is

C(5,0) = 5!/(0!)(5!) = 1

the element a alone is a subset of M

so is the element e, and i etc

so that is 5 more

in fact the number of combinations of 5 elements taken 1 at a time is

C(5,1) = 5!/(1!)(4)! = 5

ae. ai ao au are as well and ei eo eu + io iu and ou

so that is 4+3+2=1 = 10

in fact the number of combinations of 5 elements taken 2 at a time is

C(5,2) = 5!/(2!)(3!) = 10

then 5 elements taken 3 at a time

C(5,3)= 5!/(3!)(2!) =10

then 5 elements taken 4 at a time

C(5,4)= 5!/(4!)(1!) = 5

Then of course there is the 5 elements taken five at a time, the full set

C(5,5) = 5!/(5!)(0!) = 1

so we have

1 + 5 + 10 +10 + 5 + 1 or 32 subsets including the empty set and the full set.

If this looks a bit like row 5 of Pascal's triangle, that is not a coincidence. - Maths - Sets -
**Veronica**, Sunday, May 4, 2008 at 10:37amThankyou , but my book says that subsets are the whole elements in a set

how is thaT ? - Maths - Sets -
**Damon**, Sunday, May 4, 2008 at 10:43amBeats me Veronica. I would have to read the chapter in your book.

I think subsets are all the combinations of whole elements in the set.

If you paint all five vowels in a box on a piece of paper, the subsets are all the combinations of them that you can circle with a pen.

I will google that definition you have and see what I can make of it.- Maths - Sets -
**Damon**, Sunday, May 4, 2008 at 10:46amIf your class has no background in the ideas presented here, that’s fine; just start a few steps back. Review what a set is (simply "a collection of objects" will do) and what a subset is (any set of elements that all come from the original set, including Ø--the empty set--and the whole set). Make sure students understand that in a description of a set, the order of the elements does not matter (i.e., the sets {a, b} and {b, a} are the same). This will be a potential cause for confusion later in the project, since the order the subsets are listed will matter, but the order of objects listed within a subset does not.

Ask students how many subsets there are of the set {a, b, c}, and let them work on that question for quite a while. Students without a background in combinatorics will likely try to come up with a strategy for listing all the subsets, which is fine, since creating a particular kind of listing is the point of this project. Give students plenty of time to work on this problem, and ask them to convince themselves that they have not left off any subsets or listed any twice. Finally, ask several students to share their strategies with the whole class.

You may find that students will omit Ø and the complete set {a, b, c} in their listing. You can explain that Ø is a subset of any set at all, and that the whole set indeed fits the definition of subset. (If students resist calling the whole set a subset of itself, you may want to introduce the term "proper subset"; explain that it is an important distinction they are making, and one that mathematicians care about- Maths - Sets -
**Damon**, Sunday, May 4, 2008 at 10:48amThat is from

http://www2.edc.org/makingmath/mathprojects/subsets/subsets_teach.asp

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**Damon**, Sunday, May 4, 2008 at 11:09amOIC, well a is an element of the set of vowels but I suppose so is ae or aei or aeio or aeiou or any other whole group that you can draw a circle around.

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**Reiny**, Sunday, May 4, 2008 at 11:11amIn general if you have n elements, then the number of subsets is 2^n , which includes the null set and the complete set.

I you look at Damon's first reply you will notice that the numbers he has are the elements of a row in Pascal's triangle.

in any row of Pascal's triangle the sum of the entries are powers of 2,

in this case there were 5 elements, so the total would be 2^5 or 32 just as Damon showed it in detail