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Question Number 145467 by puissant last updated on 05/Jul/21

	Answered by Olaf_Thorendsen last updated on 05/Jul/21

	$a)$ $Pour former une partie B de E a k$ $elements, qui contienne un seul$ $elements de A, il faut :$ $- choisir un element de A, ce qui se$ $fait de p facons .$ $- puis choisir k - 1 elements restants$ $de B parmi les n - p elements restants$ $de E, ce qui se fait de C_{n - p}^{k - 1} facons .$ $Pour former B, il y a donc p C_{n - p}^{k - 1} facons .$ $b)$ $- B contient un element de A de$ $p C_{n - p}^{k - 1} facons ou encore C_{p}^{1} C_{n - p}^{k - 1} facons .$ $- B contient 2 elements de A de$ $C_{p}^{2} C_{n - p}^{k - 2} facons .$ $- B contient 3 elements de A de$ $C_{p}^{3} C_{n - p}^{k - 3} facons .$ $e t c . . .$ $- B contient k elements de A de$ $C_{p}^{k} C_{n - p}^{0} facons .$ $Pour repondre a la question, il suffit$ $de sommer :$ $\underset{i = 1}{\sum^{k}} C_{p}^{i} C_{n - p}^{k - i} = \underset{i = 1}{\sum^{k}} \frac{p!}{i! (p - i)!} \times \frac{(n - p)!}{(k - i)! (n - p - k + i)!}$ $\underset{i = 1}{\sum^{k}} C_{p}^{i} C_{n - p}^{k - i} = p! (n - p)! \underset{i = 1}{\sum^{k}} \frac{1}{i! (k - i)! (p - i)! (n - p - k + i)!}$
	Commented by puissant last updated on 05/Jul/21

	$merci$
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