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use permutation and combinationformulas when needed 40. Nine cards are numbeleu

ID: 3148306 • Letter: U

Question

use permutation and combinationformulas when needed 40. Nine cards are numbeleu ll I to 9 hand is dealt, 1 card at a time. How manyand,O 9. A 3 in which (A) Order is taken into consideration? 2! (B) Order is not taken into consideration? 41. Discuss the relative growth rates of x, 3', and 42. Discuss the relative growth rates of at, 2, and 10! 43. From a standard 52-card deck, how man consist entirely of red cards? consist entirely of clubs? consist entirely of face cards? consist entirely of queens? contain four kings? consist of 3 hearts and 4 diamonds? Ho 44. From a standard 52-card deck, how many 6 45. From a standard 52-card deck, how man 46 From a standard 52-card deck, how many 5-card hands 47. From a standard 52-card deck, how many 7-card hands 48. From a standard 52-card deck, how many 7-card hands 49.) From a standard 52-card deck, how many 4-card hands 50 From a standard 52-card deck, how many 4-card hands 51. A catering service offers 8 appetizers, 10 main courses, and 62 ssuming that n is an contain a card from each suit? 64. In 65. A consist of cards from the same suit? ho election to be a ain your reasoning ers: a vice-president affairs, and a 7 desserts. A banquet committee selects 3 appetizers, 4 main courses, and 2 desserts. How many ways can this be done? 52. Three departments have 12, 15, and 18 members, respec- tively. If each department selects a delegate and an altermnate to represent the department at a conference, how many ways can this be done? r vice-presidents In Problems 53 and 54, refer to the table in the graphing rary display below, which shows y1 = nPr and y,-,G for n = 1 for his mother, 1 rother 30 plate, vanilla, or 20 60 20 720

Explanation / Answer

45.
There are total 3*4 = 12 face cards in a deck of 52 cards.
5 cards can be selected in 12C5 = 792


47.
Four kings can be selected in only 1 way
remaining 3 cards from 48 cards can be selected in 48C3 = 17296 ways

49.
4 cards from each suit can be selected in 13C1*13C1*13C1*13C1 = 28561 ways

50.
4 cards from same suit = 4C1*13C4 = 2860 ways