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?


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Question Number 111394 by Aina Samuel Temidayo last updated on 03/Sep/20

$A teacher conducts a test for five$ $students . He provides the marking$ $scheme and asked them to exchange$ $their scripts such that none of them$ $marks his own script . How many ways$ $can the students carry out the$ $marking ?$
Commented by mr W last updated on 03/Sep/20

$44$
Commented by Aina Samuel Temidayo last updated on 03/Sep/20

$Solution please ?$
	Answered by mr W last updated on 03/Sep/20

	$t o t a l : 5!$ $a t l e a s t o n e s t u d e n t m a r k s h i s o w n s c r i p t :$ $C_{1}^{5} \times 4!$ $a t l e a s t t w o s t u d e n t s m a r k t h e i r o w n s c r i p t s :$ $C_{2}^{5} \times 3!$ $. . .$ $\Rightarrow 5! - C_{1}^{5} \times 4! + C_{2}^{5} \times 3! - C_{3}^{5} \times 2! + C_{4}^{5} \times 1! - 1$ $= 120 - 120 + 60 - 20 + 5 - 1$ $= 44$
	Commented by Aina Samuel Temidayo last updated on 03/Sep/20

	$General formula for! n ?$
	Commented by mr W last updated on 03/Sep/20

	$o r$ $! 5 = 44$
	Commented by mr W last updated on 03/Sep/20

	Commented by mr W last updated on 03/Sep/20

	$a s s h o w n a b o v e$ $! n = n! \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{k!}$ $w e h a v e$ $! n = (n - 1) [! (n - 1) +! (n - 2)]$ $! n = [\frac{n!}{e}]$
	Commented by mr W last updated on 03/Sep/20

	$w e s e e$ $\frac{! n}{n!} = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k}}{k!}$ $\lim_{n \to \infty} \frac{! n}{n!} = \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k}}{k!} = \frac{1}{e}$
	Commented by Aina Samuel Temidayo last updated on 04/Sep/20

	$Thanks .$
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