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Mathematics 8 Online
OpenStudy (anonymous):

what are the 32 subsets of C={m,n,o,p,q}

OpenStudy (anonymous):

a subset is a smaller grouping of the original so {m,n} {n,o} {o,p} {p,q} {m,n,o,p,q} etc. Can you get that from here?

OpenStudy (anonymous):

{m} {n} {o} {p} {q} {m,o} {p,o} {m,p} are some more...you can also put groupings of three and four of the letters together... Got it? :)

OpenStudy (anonymous):

There are 5 different elements, with the largest possible subset being 5 elements (the set C itself), and the smallest being 1 element. 5 element sets: eg {m, n, o, p, q} (5 * 4 * 3 * 2 * 1) / 5! = 1 + 4 element sets: eg, {m, n, o, p} (5 * 4 * 3 * 2) / 4! = 5 + 3 element sets: eg, {m, n, o} (5 * 4 * 3) / 3! = 10 + 2 element sets: eg, {m, n} (5 * 4) / 2! = 10 + 1 element sets: eg, {m} (5) / 1! = 5 = 31 possible subsets. Oops?

OpenStudy (zarkon):

+ the empty set \(\emptyset\)

OpenStudy (anonymous):

Here they are {{}, {m}, {n}, {o}, {p}, {q}, {m, n}, {m, o}, {m, p}, {m, q}, {n, o}, {n, p}, {n, q}, {o, p}, {o, q}, {p, q}, {m, n, o}, {m, n, p}, {m, n, q}, {m, o, p}, {m, o, q}, {m, p, q}, {n, o, p}, {n, o, q}, {n, p, q}, {o, p, q}, {m, n, o, p}, {m, n, o, q}, {m, n, p, q}, {m, o, p, q}, {n, o, p, q}, {m, n, o, p, q}}

OpenStudy (anonymous):

Why did you just give her the entire answer? We kind of discourage that on OpenStudy.

OpenStudy (anonymous):

Because, she already had all of the subsets with all the posts above. This was a summary.

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