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OpenStudy (anonymous):
How many subsets of at least four elements does a set of seven elements have?
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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"
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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
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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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