Quiz — Dénombrements divers, I.pdf

Question 1 sur 5 10:00

[{"id":443,"question":"Combien d'arrangements différents peut-on former avec 4 lettres distinctes prises 2 à 2 ?","option_a":"6","option_b":"12","option_c":"24","option_d":"4","option_e":"","option_f":"","bonne_reponse":"c","explication":"Le nombre d'arrangements de 4 éléments pris 2 à 2 est donné par la formule A(4,2) = 4! \/ (4-2)! = 4 × 3 = 12.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"c\", \"options\": {\"a\": \"6\", \"b\": \"12\", \"c\": \"24\", \"d\": \"4\"}}","_debug_options_count":4},{"id":444,"question":"Quel est le coefficient binomial C(5,3) ?","option_a":"10","option_b":"15","option_c":"20","option_d":"5","option_e":"","option_f":"","bonne_reponse":"a","explication":"C(5,3) = 5! \/ (3! × (5-3)!) = (5×4) \/ (2×1) = 10. Ce résultat correspond au nombre de combinaisons de 5 éléments pris 3 à 3.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"10\", \"b\": \"15\", \"c\": \"20\", \"d\": \"5\"}}","_debug_options_count":4},{"id":445,"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":"4","option_e":"","option_f":"","bonne_reponse":"b","explication":"Le nombre de permutations de 4 lettres distinctes est 4! = 4 × 3 × 2 × 1 = 24.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"12\", \"b\": \"24\", \"c\": \"6\", \"d\": \"4\"}}","_debug_options_count":4},{"id":446,"question":"Dans un groupe de 6 personnes, combien de façons peut-on choisir un comité de 3 membres ?","option_a":"20","option_b":"18","option_c":"15","option_d":"12","option_e":"","option_f":"","bonne_reponse":"a","explication":"Il s'agit d'une combinaison : C(6,3) = 6! \/ (3! × 3!) = 20. L'ordre n'intervient pas dans le choix des membres.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"20\", \"b\": \"18\", \"c\": \"15\", \"d\": \"12\"}}","_debug_options_count":4},{"id":447,"question":"Quel est le nombre de mots de 3 lettres (même si elles ne sont pas des mots valides) que l'on peut former avec l'alphabet de 26 lettres, sans répétition ?","option_a":"15 600","option_b":"17 550","option_c":"26 000","option_d":"12 500","option_e":"","option_f":"","bonne_reponse":"a","explication":"C'est un arrangement de 26 lettres prises 3 à 3 : A(26,3) = 26! \/ (26-3)! = 26 × 25 × 24 = 15 600.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"15 600\", \"b\": \"17 550\", \"c\": \"26 000\", \"d\": \"12 500\"}}","_debug_options_count":4}]