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Question Number 111394 by Aina Samuel Temidayo last updated on 03/Sep/20
A teacher conducts a test for five  students. He provides the marking  scheme and asked them to exchange  their scripts such that none of them  marks his own script. How many ways  can the students carry out the  marking?
$$\mathrm{A}\:\mathrm{teacher}\:\mathrm{conducts}\:\mathrm{a}\:\mathrm{test}\:\mathrm{for}\:\mathrm{five} \\ $$$$\mathrm{students}.\:\mathrm{He}\:\mathrm{provides}\:\mathrm{the}\:\mathrm{marking} \\ $$$$\mathrm{scheme}\:\mathrm{and}\:\mathrm{asked}\:\mathrm{them}\:\mathrm{to}\:\mathrm{exchange} \\ $$$$\mathrm{their}\:\mathrm{scripts}\:\mathrm{such}\:\mathrm{that}\:\mathrm{none}\:\mathrm{of}\:\mathrm{them} \\ $$$$\mathrm{marks}\:\mathrm{his}\:\mathrm{own}\:\mathrm{script}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{ways} \\ $$$$\mathrm{can}\:\mathrm{the}\:\mathrm{students}\:\mathrm{carry}\:\mathrm{out}\:\mathrm{the} \\ $$$$\mathrm{marking}? \\ $$
Commented by mr W last updated on 03/Sep/20
44
$$\mathrm{44} \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
Solution please?
$$\mathrm{Solution}\:\mathrm{please}? \\ $$
Answered by mr W last updated on 03/Sep/20
total: 5!  at least one student marks his own script:  C_1 ^5 ×4!  at least two students mark their own scripts:  C_2 ^5 ×3!  ...  ⇒5!−C_1 ^5 ×4!+C_2 ^5 ×3!−C_3 ^5 ×2!+C_4 ^5 ×1!−1  =120−120+60−20+5−1  =44
$${total}:\:\mathrm{5}! \\ $$$${at}\:{least}\:{one}\:{student}\:{marks}\:{his}\:{own}\:{script}: \\ $$$${C}_{\mathrm{1}} ^{\mathrm{5}} ×\mathrm{4}! \\ $$$${at}\:{least}\:{two}\:{students}\:{mark}\:{their}\:{own}\:{scripts}: \\ $$$${C}_{\mathrm{2}} ^{\mathrm{5}} ×\mathrm{3}! \\ $$$$… \\ $$$$\Rightarrow\mathrm{5}!−{C}_{\mathrm{1}} ^{\mathrm{5}} ×\mathrm{4}!+{C}_{\mathrm{2}} ^{\mathrm{5}} ×\mathrm{3}!−{C}_{\mathrm{3}} ^{\mathrm{5}} ×\mathrm{2}!+{C}_{\mathrm{4}} ^{\mathrm{5}} ×\mathrm{1}!−\mathrm{1} \\ $$$$=\mathrm{120}−\mathrm{120}+\mathrm{60}−\mathrm{20}+\mathrm{5}−\mathrm{1} \\ $$$$=\mathrm{44} \\ $$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20
General formula for !n ?
$$\mathrm{General}\:\mathrm{formula}\:\mathrm{for}\:!\mathrm{n}\:? \\ $$
Commented by mr W last updated on 03/Sep/20
or  !5=44
$${or} \\ $$$$!\mathrm{5}=\mathrm{44} \\ $$
Commented by mr W last updated on 03/Sep/20
Commented by mr W last updated on 03/Sep/20
as shown above  !n=n!Σ_(k=0) ^n (((−1)^k )/(k!))  we have  !n=(n−1)[!(n−1)+!(n−2)]  !n=[((n!)/e)]
$${as}\:{shown}\:{above} \\ $$$$!{n}={n}!\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!} \\ $$$${we}\:{have} \\ $$$$!{n}=\left({n}−\mathrm{1}\right)\left[!\left({n}−\mathrm{1}\right)+!\left({n}−\mathrm{2}\right)\right] \\ $$$$!{n}=\left[\frac{{n}!}{{e}}\right] \\ $$
Commented by mr W last updated on 03/Sep/20
we see  ((!n)/(n!))=Σ_(k=0) ^n (((−1)^k )/(k!))  lim_(n→∞) ((!n)/(n!))=Σ_(k=0) ^∞ (((−1)^k )/(k!))=(1/e)
$${we}\:{see} \\ $$$$\frac{!{n}}{{n}!}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{!{n}}{{n}!}=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!}=\frac{\mathrm{1}}{{e}} \\ $$
Commented by Aina Samuel Temidayo last updated on 04/Sep/20
Thanks.
$$\mathrm{Thanks}. \\ $$

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