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Chapter 7: Problem 395

Two cards are to be drawn in order from a pack of 4 cards (say, an ace, king, queen, and jack), the drawn card not being replaced before the second card is drawn. How many different drawings are possible?

Short Answer

Expert verified
There are 12 different ways to draw 2 cards in a specific order from a pack of 4 cards without replacement, using the formula for permutations: \(4P2 = \frac{4!}{(4-2)!} = \frac{24}{2} = 12\).

Step by step solution

01

Analyze the given problem

We are given a pack of 4 cards: ace, king, queen, and jack. We need to find the total number of ways to draw 2 cards from this pack in a specific order, without replacement. Since the order matters and there is no replacement, we can use permutations to solve this problem.
02

Use the formula for permutations

The formula for permutations is: \[ nPr = \frac{n!}{(n-r)!} \] Where 'n' represents the total number of objects and 'r' represents the number of objects you want to choose from 'n' objects in a particular order, the exclamation point, "!", indicates a factorial function. In our case, we have 4 cards to choose from (n=4) and we want to draw 2 cards (r=2). So, we have to find 4P2 using the formula.
03

Calculate the permutation

Now, we'll calculate the permutation using the formula: \[ 4P2 = \frac{4!}{(4-2)!} \] First, we need to find the factorials of 4 and 2: 4! = 4 x 3 x 2 x 1 = 24 2! = 2 x 1 = 2 Now, plug the factorial values into the formula: \[ 4P2 = \frac{24}{2} \]
04

Final calculation

Now, divide 24 by 2 to get the final answer: \[ 4P2 = \frac{24}{2} = 12 \] So, there are 12 different ways to draw 2 cards in a specific order from a pack of 4 cards without replacement.

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