Question

lesson 1 practice. fundamental counting principle 1. a person has two pairs of shoes (black and blue), and three different pairs of socks (white, green, and yellow). determine the maximum number of sock - shoe combinations, using: a) the fundamental counting principle \\(1\cdot\frac{2}{1}\cdot\frac{3\cdot3\cdot1}{1}\\) \\(2 + 6 = 8\\) 2. in track and field, there are eight competitors in the 100 metre sprint final. in how many ways can the gold, silver, and bronze be selected?

Explanation:

Problem 1:

Step1: Identify the number of choices for each category

We have 2 pairs of shoes (black and blue) and 3 pairs of socks (white, green, and yellow). So, the number of shoe choices \( n_{shoes} = 2 \) and the number of sock choices \( n_{socks} = 3 \).

Step2: Apply the fundamental counting principle

The fundamental counting principle states that if there are \( m \) ways to do one thing and \( n \) ways to do another thing, then there are \( m\times n \) ways to do both things. Here, the number of sock - shoe combinations is the product of the number of shoe choices and the number of sock choices.
So, the number of combinations \( N=n_{shoes}\times n_{socks}\)
Substitute \( n_{shoes} = 2 \) and \( n_{socks}=3 \) into the formula: \( N = 2\times3=6 \)

Step1: Identify the problem type

This is a permutation problem because the order in which we select the gold, silver, and bronze medalists matters (gold is different from silver, etc.). We have 8 competitors and we want to select 3 (for gold, silver, bronze) and arrange them.

Step2: Use the permutation formula

The permutation formula is \( P(n,r)=\frac{n!}{(n - r)!}\), where \( n = 8 \) (total number of competitors) and \( r=3 \) (number of medals to be awarded: gold, silver, bronze).

First, calculate \( n!=8! = 8\times7\times6\times5\times4\times3\times2\times1\)
and \( (n - r)!=(8 - 3)!=5!=5\times4\times3\times2\times1\)

Then \( P(8,3)=\frac{8!}{(8 - 3)!}=\frac{8!}{5!}=\frac{8\times7\times6\times5!}{5!}\)

The \( 5! \) terms in the numerator and denominator cancel out, leaving \( 8\times7\times6\)

Step3: Calculate the result

\( 8\times7\times6 = 336 \)

Answer:

The maximum number of sock - shoe combinations is 6.

Problem 2: