1A. You need to have a password with 5 letters followed by 3 odd digits between

Question

1A. You need to have a password with 5 letters followed by 3 odd digits between 0 and 9, inclusive. If the characters and digits cannot be used more than once, how many choices do you have for your password?

1B. 5 cards are drawn from a standard deck of 52 playing cards. How many different 5-card hands are possible if the drawing is done without replacement?

1C. The newly elected president needs to decide the remaining 8 spots available in the cabinet he/she is appointing. If there are 10 eligible candidates for these positions (where rank matters), how many different ways can the members of the cabinet be appointed?

1D. You are ordering a hamburger and can get up to 7 toppings, but each topping can only be used once. You tell the cashier to surprise you with the toppings you get. What is the probability that you get 4 toppings? Express your answer as a fraction or a decimal number rounded to four decimal places.

Explanation / Answer

a) Number of choices for 5 letters = 26P5 = 26! / (26-5)! = 7893600

Number of choices for 3 odd digits = 5P3 = 5! / (5-3)! = 20

Total numbe rof choices = 7893600 * 20 = 157872000

b) Number of different 5 card hands possible = 52C5 = 52! / (5! * (52-5)!) = 2598960

c) If order of selection is important, we will use permutation.

Number of different ways the members can be appointed = 10P8 = 10! / (10-8)! = 90

d) Total number of wasy of getting toppings = 7C1 + 7C2 + 7C3 + 7C4 + 7C5 + 7C6 + 7C8 = 27 - 1 = 127

P(getting 4 toppings) = 7C4 / 127 = 35/127 = 0.2756

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