S1 Question (again sorry)

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Sang
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Report Thread starter 16 years ago
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Ok if you havent guessed im learning S1 on my own (OCR) so thats why I have a few questions.

Ive got to the Permutations and Combinations Chapter and dont understand it that much all the questions seem to be riddles?? Are all calculations worked out using the nPr and nCr buttons or are there any formulas that help answer the questions?

I have an exam question here (no answers) I dont know where to begin, or how to layout my workings.

________________________________ ________________________________ _

A standard pack of playing cards consists of 52 distinct cards. Five different cards are selected at random. The order in which the cards are selected does not matter.

(a) Find the number of different possible selections of 5 cards. [1]

(b) There are 4 suits and each suit consists of 13 cards. How many of the selections in part (a) consist of 3 spades and 2 clubs? [2]

(c)How many of the selections in part (a) contain exactly 3 spades? [2]

(d) Calculate the probability that 5 cards selected at random will consist of 3 spades and 2 clubs. [2]


I take it part (a) is simply 5! but then onwards im lost any help? Also with the general technique to answering these questions and not just this example - Much Appreciate any help UKL!!!!
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Jonny W
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Report 16 years ago
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(a) 52!/(5! 47!) = 52C5. (Lay out the 52 cards in a row, then disregard the order of the first 5 and the last 47. Or ... the number of ordered selections of 5 cards = 52*51*50*49*48, since there are 52 possible choices of the first card, 51 of the second, and so on. Then the number of unordered selections of 5 cards = number of ordered selections / 5!, since each unordered selection can be ordered in 5! ways.)

(b) 13C3 * 13C2. (There are 13C3 ways of choosing 3 spades, and 13C2 ways of choosing 2 clubs.)

(c) 13C3 * 39C2. (There are 13C3 ways of choosing 3 spades, and 39C2 ways of choosing 2 non-spades.)

(d) Ans to (b) / Ans to (a).
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