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there-are-100-students-in-a-school-it-is-found-out-that-each-student-should-select-at-least-4-courses-so-that-no-two-students-have-the-same-selection-how-many-different-courses-does-the-school




Question Number 218445 by mr W last updated on 10/Apr/25
there are 100 students in a school.  it is found out that each student   should select at least 4 courses, so   that no two students have the same   selection.   how many different courses does   the school offer?
$${there}\:{are}\:\mathrm{100}\:{students}\:{in}\:{a}\:{school}. \\ $$$${it}\:{is}\:{found}\:{out}\:{that}\:{each}\:{student}\: \\ $$$${should}\:{select}\:{at}\:{least}\:\mathrm{4}\:{courses},\:{so}\: \\ $$$${that}\:{no}\:{two}\:{students}\:{have}\:{the}\:{same}\: \\ $$$${selection}.\: \\ $$$${how}\:{many}\:{different}\:{courses}\:{does}\: \\ $$$${the}\:{school}\:{offer}? \\ $$
Answered by MrGaster last updated on 10/Apr/25
C:∪_(k=4) ^v  ((([v])),(k) )⊇{S_i }_(i=1) ^(100) ,S_i ⊆[v],∣S_i ∣≥4,S_i ≠S_j ∀i≠j  min v s.t∣∪_(k=4) ^v  ((([v])),(k) )∣≥100  ⇒Σ_(k=4) ^v  ((v),(k) )≥100  Lower bound via  ((v),(4) )≤Σ_(k=4) ^v  ((v),(4) )<2^v    ((v),(4) )=((v(v−1)(v−2)(v−3))/(24))≥100  ⇒v(v−1)(v−2)(v−3)≥2400  Test v=9: 9×8×7×6=3024≥2400  Test v=8: 8×7×6×5=1680<2400  ⇒ determinant ((9))
$$\mathbb{C}:\underset{{k}=\mathrm{4}} {\overset{{v}} {\cup}}\begin{pmatrix}{\left[{v}\right]}\\{{k}}\end{pmatrix}\supseteq\left\{\mathcal{S}_{{i}} \right\}_{{i}=\mathrm{1}} ^{\mathrm{100}} ,\mathcal{S}_{{i}} \subseteq\left[{v}\right],\mid\mathcal{S}_{{i}} \mid\geq\mathrm{4},\mathcal{S}_{{i}} \neq\mathcal{S}_{{j}} \forall{i}\neq{j} \\ $$$$\mathrm{min}\:{v}\:\mathrm{s}.\mathrm{t}\mid\underset{{k}=\mathrm{4}} {\overset{{v}} {\cup}}\begin{pmatrix}{\left[{v}\right]}\\{{k}}\end{pmatrix}\mid\geq\mathrm{100} \\ $$$$\Rightarrow\underset{{k}=\mathrm{4}} {\overset{{v}} {\sum}}\begin{pmatrix}{{v}}\\{{k}}\end{pmatrix}\geq\mathrm{100} \\ $$$$\mathrm{Lower}\:\mathrm{bound}\:\mathrm{via}\:\begin{pmatrix}{{v}}\\{\mathrm{4}}\end{pmatrix}\leq\underset{{k}=\mathrm{4}} {\overset{{v}} {\sum}}\begin{pmatrix}{{v}}\\{\mathrm{4}}\end{pmatrix}<\mathrm{2}^{{v}} \\ $$$$\begin{pmatrix}{{v}}\\{\mathrm{4}}\end{pmatrix}=\frac{{v}\left({v}−\mathrm{1}\right)\left({v}−\mathrm{2}\right)\left({v}−\mathrm{3}\right)}{\mathrm{24}}\geq\mathrm{100} \\ $$$$\Rightarrow{v}\left({v}−\mathrm{1}\right)\left({v}−\mathrm{2}\right)\left({v}−\mathrm{3}\right)\geq\mathrm{2400} \\ $$$$\mathrm{Test}\:{v}=\mathrm{9}:\:\mathrm{9}×\mathrm{8}×\mathrm{7}×\mathrm{6}=\mathrm{3024}\geq\mathrm{2400} \\ $$$$\mathrm{Test}\:{v}=\mathrm{8}:\:\mathrm{8}×\mathrm{7}×\mathrm{6}×\mathrm{5}=\mathrm{1680}<\mathrm{2400} \\ $$$$\Rightarrow\begin{array}{|c|}{\mathrm{9}}\\\hline\end{array} \\ $$
Commented by mr W last updated on 10/Apr/25
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Answered by mr W last updated on 10/Apr/25
say the school offers n courses.  C_n ^4 ≥100 ∧ C_n ^3 <100  ⇒n=9
$${say}\:{the}\:{school}\:{offers}\:{n}\:{courses}. \\ $$$${C}_{{n}} ^{\mathrm{4}} \geqslant\mathrm{100}\:\wedge\:{C}_{{n}} ^{\mathrm{3}} <\mathrm{100} \\ $$$$\Rightarrow{n}=\mathrm{9} \\ $$
Answered by Spillover last updated on 12/Apr/25
n!/4!/(n-4)!=100  n!/(n-4)!=2400  n(n-1)(n-2)(n-3)=2400  8<n<9  N=9
n!/4!/(n-4)!=100
n!/(n-4)!=2400
n(n-1)(n-2)(n-3)=2400
8<n<9
N=9

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