Mathematics algebra1 algebra2

Question

Calculate the number of subsets and the number of proper subsets for the set

{1,3,5,7,9,11}

Answer 1

There are 2⁶ - 1 proper subsets and 2⁶ subsets, including the null set.

The number of ways of choosing groups (subsets) from 6 things is the sum of

₆C_n from n = 0 to 6 which is 2⁶ including the null set.

Answer 2

For this question, ask yourself what is really important? What information do I need to solve the problem, and what is just extra? It is only asking you to count the number of subsets, so you can ignore what the numbers themselves and just focus on the fact that it has 6 unique items. So take a moment and ask yourself, how many ways can choose groups from 6 items? You should see if you can figure it out this way bfore reading further.

OK, so how many ways are there? I'm going to just use a set of 4 letters here to make this simpler.

First, let's choose each individual item: a, b, c, d = 4.

Next, let's choose two at a time: ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc = 12, but we don't care about order (because the question didn't say ordered subsets), so we can eliminate all the duplicates to give us 6.

Next, let's choose three at a time (and just ignore duplicates): abc, abd, bcd = 3

And finally all 4: abcd = 1

If you look at the pattern, what you are doing is finding the combinations, which you can express as n(n-1)(n-2) . . . k times. This looks a lot like a factorial n!, except that you stop after k times. You can express this n!/(n-k)!

If that part doesn't make sense, pause and really think about it. Play around with a few sets of different sizes. Think about writing n! as n(n-1)(n-2) . . . (n-k)(n-k-1) . . . 1

But wait, that still gives us all of those dupicates we don't care about. Consider the combinations abc, acb, bac, bca, cab, cba. For each starting item, the number of possible ways to add the others is just another combination. So you wind up with k(k-1)(k-2) . . . 1 copies of the unique sets. That's just k!

So you can choose k items from a set of n items in n!/k!(n-k)! ways. This formula is called the "n choose k" formula, and it's so important that it has it's own denotation that looks like a fraction inside a parentheses, but with no fraction line.

I'd recommend pausing again and thinking about how you can use this to solve your problem before reading on.

So back to your problem: for subsets, we have:

(6 choose 6) + (6 choose 5) + (6 choose 4) + (6 choose 3) + (6 choose 2) + (6 choose 1)

Scientific and graphing calculators have this function built in, so all you need to do to solve your problem is set it up and put it in the calculator. Google calculator also works by just writing out n choose k (with n and k replaced with your numbers).

For the second part of the question, you want all of these subsets except the one that is equal to the original, so you just get rid of (6 choose 6) or just take the results from the first part and subtract 1.