Dénombrements : Permutations, Arrangements et Combinaisons — Terminale Math

Question 1 sur 10 10:00

[{"id":30288,"question":"Combien de permutations différentes peut-on former avec les lettres du mot 'MATH' ?","option_a":"12","option_b":"24","option_c":"6","option_d":"16","option_e":"","option_f":"","bonne_reponse":"B","explication":"Il y a 4 lettres distinctes, donc 4! = 24 permutations possibles.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30289,"question":"Vrai ou Faux : Le nombre d'arrangements de 3 éléments parmi 5 est égal à C(5,3).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"C(5,3) donne le nombre de combinaisons (sans ordre), tandis que les arrangements A(5,3) = 5!\/(5-3)! = 60 tiennent compte de l'ordre.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30290,"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":30291,"question":"Vrai ou Faux : Le principe multiplicatif s'applique uniquement aux arrangements.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"Le principe multiplicatif s'applique à tous les types de dénombrement (permutations, arrangements, combinaisons) pour compter des configurations.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30292,"question":"Combien de nombres à 3 chiffres distincts peut-on former avec les chiffres 1, 2, 3, 4 et 5 ?","option_a":"60","option_b":"20","option_c":"40","option_d":"120","option_e":"","option_f":"","bonne_reponse":"A","explication":"Il y a 5 choix pour le premier chiffre, 4 pour le second et 3 pour le troisième : 5 × 4 × 3 = 60.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30293,"question":"Vrai ou Faux : A(n,k) = n! \/ (n-k)!.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"A(n,k) = n! \/ (n-k)! est la formule des arrangements, mais elle est correcte.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30294,"question":"Quel est le développement de (a + b)^3 selon le binôme de Newton ?","option_a":"a³ + 3a²b + 3ab² + b³","option_b":"a³ + 2a²b + 2ab² + b³","option_c":"a³ + b³","option_d":"a³ + 3ab + 3a²b² + b³","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le binôme de Newton donne (a+b)^3 = C(3,0)a³ + C(3,1)a²b + C(3,2)ab² + C(3,3)b³ = a³ + 3a²b + 3ab² + b³.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30295,"question":"Vrai ou Faux : C(n,k) = C(n,n-k).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"C'est une propriété fondamentale des coefficients binomiaux : choisir k éléments parmi n revient au même que d'en laisser n-k de côté.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30296,"question":"Combien de comités de 4 personnes peut-on former à partir d'un groupe de 10 personnes ?","option_a":"210","option_b":"120","option_c":"240","option_d":"360","option_e":"","option_f":"","bonne_reponse":"A","explication":"C(10,4) = 10! \/ (4! × 6!) = 210 combinaisons possibles.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30297,"question":"Vrai ou Faux : Le nombre de permutations d'un ensemble de n éléments est toujours un multiple de n.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Le nombre de permutations est n!, qui est toujours divisible par n (car n! = n × (n-1)!).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]