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Mathematics 21 Online
OpenStudy (anonymous):

How many subsets of at least four elements does a set of seven elements have?

OpenStudy (anonymous):

\(\left(\begin{matrix}7 \\ 4\end{matrix}\right)\)I think.

OpenStudy (anonymous):

7 choose four should work... That 7factorial(6factorial)....(1factorial) divided by 3factorial by four factorial

OpenStudy (anonymous):

7x6x5 x 4 / 4x3x2x1 = 35

OpenStudy (anonymous):

bendt has it \[\dbinom{7}{4}\] \[=\frac{7\times 6\times 5}{3\times 2}=7\times 5=35\]

OpenStudy (amistre64):

at least 4; not just "4"

OpenStudy (anonymous):

oh no that is not right!

OpenStudy (anonymous):

"at least 4!"

OpenStudy (amistre64):

7c4 + 7c5 + 7c6 + 7c7 right?

OpenStudy (anonymous):

so 1 or 2 or 3 or 4. we just did number of subsets with 4 elements. need the others as well. amistre is right

OpenStudy (amistre64):

3560 out of 12819 XP done lol

OpenStudy (anonymous):

\[\dbinom{7}{5}=\frac{7\times 6}{2}=21\] \[d\binom{7}{6}=7\] and \[\dbinom{7}{7}=1\] so answer is \[1+7+21+35\]

OpenStudy (anonymous):

what does 3560 out of 12819XP mean??

OpenStudy (blacksteel):

Yeah, it's going to be 7c4 + 7c5 + 7c6 + 7c7, or \[7!/(4!*3!) + 7!/(5!*2!) + 7!/(6!*1!) + 7!/7!\] or 35 + 21 + 7 + 1 = 64.

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