Questions: Counting and Probability Introduction to permutations and combinations Suppose we want to choose 2 objects, without replacement, from the 4 objects pencil, eraser, desk, and chair (a) How many ways can this be done, if the order of the choices is relevant? (b) How many ways can this be done, if the order of the choices is not relevant?

Counting and Probability
Introduction to permutations and combinations

Suppose we want to choose 2 objects, without replacement, from the 4 objects pencil, eraser, desk, and chair
(a) How many ways can this be done, if the order of the choices is relevant?

(b) How many ways can this be done, if the order of the choices is not relevant?
Transcript text: Counting and Probability Introduction to permutations and combinations Suppose we want to choose 2 objects, without replacement, from the 4 objects pencil, eraser, desk, and chair (a) How many ways can this be done, if the order of the choices is relevant? (b) How many ways can this be done, if the order of the choices is not relevant?
failed

Solution

failed
failed

Solution Steps

Step 1: Identify the Problem Type

Given the order of selection is relevant, we are dealing with a permutations problem.

Step 2: Apply the Permutations Formula

The formula for permutations is given by \(P(n, k) = \frac{n!}{(n-k)!}\).

Step 3: Substitute the Values \(n = 4\) and \(k = 2\)
Calculation: \(P(4, 2) = \frac{4!}{(4-2)!} = 12\)

Final Answer: The number of ways to choose 2 objects from 4 distinct objects is 12.

Step 1: Identify the Problem Type

Given the order of selection is not relevant, we are dealing with a combinations problem.

Step 2: Apply the Combinations Formula

The formula for combinations is given by \(C(n, k) = \frac{n!}{k!(n-k)!}\).

Step 3: Substitute the Values \(n = 4\) and \(k = 2\)
Calculation: \(C(4, 2) = \frac{4!}{2!(4-2)!} = 6\)
Final Answer: The number of ways to choose 2 objects from 4 distinct objects is 6.
Was this solution helpful?
failed
Unhelpful
failed
Helpful