/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 A starting lineup in basketball ... [FREE SOLUTION] | 91Ó°ÊÓ

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A starting lineup in basketball consists of two guards, two forwards, and a center. a. A certain college team has on its roster three centers, four guards, four forwards, and one individual (X) who can play either guard or forward. How many different starting lineups can be created? b. Now suppose the roster has 5 guards, 5 forwards, 3 centers, and 2 "swing players" \((X\) and \(Y\) ) who can play either guard or forward. If 5 of the 15 players are randomly selected, what is the probability that they constitute a legitimate starting lineup?

Short Answer

Expert verified
a. 300 lineups b. Probability is approximately 43.9%.

Step by step solution

01

Understand the problem

We need to determine how many different starting lineups for a basketball team are possible given a roster with specific player arrangements and restrictions. The lineups have specific roles: two guards, two forwards, and one center.
02

Count possible lineups for Part (a)

For part (a), we have 3 centers, 4 guards, and 4 forwards. Player X can be either a guard or a forward. - Choose 1 center out of 3: \(3\) ways.- Choose 2 guards out of 5 (4 regular + X who can also be a guard): \(\binom{5}{2} = 10\) ways.- Choose 2 forwards out of 5 (4 regular + X who can also be a forward): \(\binom{5}{2} = 10\) ways.Multiply these choices: \(3 \times 10 \times 10 = 300\) possible lineups.
03

Compute probability for Part (b)

For part (b), we can select 5 players from 15, and any selection is equally probable. The total number of 5-player combinations is \(\binom{15}{5} = 3003\).Next, calculate the successful outcomes where we form a legitimate lineup:- Choose 1 center from 3 centers: \(3\) ways.- Choose 2 guards from 7 options (5 guards + X and Y): \(\binom{7}{2} = 21\) ways.- Choose 2 forwards from 7 options (5 forwards + X and Y): \(\binom{7}{2} = 21\) ways.Multiply these choices: \(3 \times 21 \times 21 = 1323\) successful outcomes.Therefore, the probability of selecting a legitimate starting lineup is: \(\frac{1323}{3003} = \frac{441}{1001}\), simplifying further gives approximately \(0.439\) or about 43.9%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability calculations
Probability calculations are often used to determine the likelihood of a specific event occurring. It involves comparing the number of favorable outcomes to the total number of possible outcomes. In the basketball lineup problem described, the task is to find the probability that randomly selected players form a legitimate starting lineup.
We first count all possible combinations of players that can be chosen. Given 15 players, there are \(inom{15}{5}\) ways to choose any 5 players. However, we're interested in legitimate lineups, which means we need a specific player combination: two guards, two forwards, and one center.
To find the probability:
  • Calculate the success combinations (legitimate lineups)
  • Divide the number of successful combinations by the total combinations
In part (b) of the problem, we found there are 1323 legitimate combinations, and since there are 3003 possible ways to randomly select 5 players, the probability of forming a valid lineup is \(\frac{1323}{3003}\). This means there is about a 43.9% chance of randomly selecting a valid lineup.
Binomial coefficient
The binomial coefficient is a key concept in combinatorics, representing the number of ways to choose a subset of elements from a larger set. It's denoted as \(\binom{n}{k}\), pronounced as 'n choose k'. It gives the count of ways to select \(k\) elements from a \(n\) element set without regard to order.
Mathematically, it is computed using the formula: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]For instance, in the lineup problem, we need to use the binomial coefficient to determine the number of ways to select players for specific positions. When choosing guards and forwards, we calculated \(\binom{5}{2}\) as there were options for both guards and forwards, including the swing players X and Y.
  • The binomial coefficient helps efficiently count combinations in complex probability problems.
  • It simplifies determining subsets in unordered selections.
Using these calculations, you can quickly determine the number of combinations for each position needed in the basketball problem.
Basketball lineup problem
The basketball lineup problem is a classic example of using combinatorial methods to solve a real-world scenario. In basketball, a starting lineup must consist of specific player roles: two guards, two forwards, and one center. The challenge is to calculate how many such lineups are possible given different player configurations.
In the given problem, we are asked to determine the number of possible lineups with a specific player roster. Different permutations and combinations need to be evaluated based on available and flexible players (like swing players X and Y) who can fit multiple roles.
  • Identify the roster's available positions
  • Use combinations to fill each position
  • Account for players who can fill multiple positions (e.g., swing players)
    • This exercise allows students to apply combinatorial logic to organize and establish clear lineup formations while understanding roles and restrictions. It's essential in understanding how variables and constraints affect combinations, providing a practical application of mathematical principles in sports.

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