Maîtrisez l'Analyse Combinatoire en 10 questions !

Question 1 sur 10 10:00

[{"id":13803,"question":"Combien de permutations existe-t-il pour les lettres du mot 'MATH' ?","option_a":"12","option_b":"24","option_c":"6","option_d":"48","option_e":"","option_f":"","bonne_reponse":"B","explication":"Le mot 'MATH' contient 4 lettres distinctes. Le nombre de permutations est donc 4! = 4 × 3 × 2 × 1 = 24.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13804,"question":"Le nombre de combinaisons de 3 éléments parmi 5 est égal à :","option_a":"10","option_b":"15","option_c":"20","option_d":"60","option_e":"","option_f":"","bonne_reponse":"A","explication":"C(5,3) = 5! \/ (3! × 2!) = (5×4)\/(2×1) = 10.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13805,"question":"Vrai ou Faux ? Le nombre d'arrangements de 2 éléments parmi 4 est égal à 12.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le nombre d'arrangements A(4,2) = 4! \/ (4-2)! = 4 × 3 = 12. L'affirmation est donc vraie.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13806,"question":"Quel est le coefficient binomial C(6,2) ?","option_a":"15","option_b":"30","option_c":"20","option_d":"12","option_e":"","option_f":"","bonne_reponse":"A","explication":"C(6,2) = 6! \/ (2! × 4!) = (6×5)\/(2×1) = 15.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13807,"question":"Vrai ou Faux ? Le principe additif s'applique lorsque les événements sont mutuellement exclusifs.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le principe additif s'utilise pour calculer la probabilité de l'union d'événements mutuellement exclusifs : P(A ∪ B) = P(A) + P(B).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13808,"question":"Combien de codes de 4 chiffres peut-on former avec les chiffres de 0 à 9 sans répétition ?","option_a":"5040","option_b":"210","option_c":"126","option_d":"3024","option_e":"","option_f":"","bonne_reponse":"A","explication":"Il s'agit d'un arrangement de 4 chiffres parmi 10 : A(10,4) = 10! \/ 6! = 10 × 9 × 8 × 7 = 5040.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13809,"question":"Vrai ou Faux ? Une combinaison de 5 éléments parmi 5 est toujours égale à 1.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"C(5,5) = 5! \/ (5! × 0!) = 1. L'affirmation est donc vraie.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13810,"question":"Quel est le résultat de 7! \/ 5!?","option_a":"42","option_b":"21","option_c":"5040","option_d":"120","option_e":"","option_f":"","bonne_reponse":"A","explication":"7! \/ 5! = (7×6×5!) \/ 5! = 7 × 6 = 42.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13811,"question":"Dans une classe de 20 élèves, combien de groupes de 3 élèves peut-on former ?","option_a":"1140","option_b":"6840","option_c":"2850","option_d":"1710","option_e":"","option_f":"","bonne_reponse":"A","explication":"C(20,3) = 20! \/ (3! × 17!) = (20×19×18)\/(3×2×1) = 1140.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":13812,"question":"Vrai ou Faux ? Le nombre de permutations de n éléments est toujours égal à n^n.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"Le nombre de permutations de n éléments est n! (factorielle de n), et non n^n. Par exemple, pour n=3, 3! = 6 ≠ 3^3 = 27.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]