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There are 15 qualified applicants for 5 trainee positions in a fast-food management program. How many different groups of trainees can be selected?

Short Answer

Expert verified
There are 3003 different groups of trainees that can be selected.

Step by step solution

01

Understanding the Problem

We need to find how many different groups of 5 trainees can be selected from a pool of 15 applicants. This is a typical problem of combinations, where the order of selection does not matter.
02

Define as a Combination

This problem can be represented as a combination because we are selecting a subset (5 trainees) from a larger set (15 applicants) without regard to order. The mathematical representation is '\( \binom{15}{5} \)'.
03

Apply the Combination Formula

The formula for combinations is given by \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. For this problem, \( n = 15 \) and \( r = 5 \).
04

Calculate Factorials

Calculate the factorials needed: \( 15! = 15 \times 14 \times 13 \times 12 \times 11 \times 10! \), \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \), and \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \).
05

Simplify the Combination Calculation

Substitute the factorials into the combination formula: \( \binom{15}{5} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \). Cancel out the common terms: \( \frac{360360}{120} \).
06

Compute the Result

Finish the calculation to get \( 360360 \div 120 = 3003 \). Thus, there are 3003 different ways to select 5 trainees from 15 applicants.

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Key Concepts

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

Combinations
When faced with problems where selection is key, like choosing a subset of trainees from a larger group, combinations become essential. In combinatorics, a combination refers to a way of selecting items from a larger set, such that the order of selection doesn’t matter. This is why, in our example with 15 applicants and 5 positions, we use combinations to find the answer.

The combination formula is denoted as \( \binom{n}{r} \), which translates to choosing \( r \) items from \( n \) options. The formula is structured as:
  • \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)
This suggests that you calculate the number of ways to combine different items by dividing the factorial of the total number by the factorials of chosen items and the difference from what you're choosing out of.

Remember, the key distinction in combination is that the arrangement in your chosen subset does not impact the outcome. Selecting trainees A, B, and C is the same as selecting them in any other order like B, C, A.
Factorials
Factorials are a fundamental concept when dealing with combinations and permutations. They help in expressing how to calculate the number of possible arrangements of a set of items. Expressed with an exclamation mark, \( n! \) (n factorial) defines the product of all positive integers up to \( n \).

For example:
  • \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
  • \( 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)
When breaking down a combination calculation, like \( \binom{15}{5} \), factorials simplify how we manage the larger numbers involved (like 15!). With simplification, many terms will cancel out, making it easier to compute the final result.

Factorials are especially useful in simplifying complex mathematical computations, as they often allow numbers to be reduced through cancellation during division.
Permutation vs Combination
Differentiating between permutation and combination is crucial in solving problems like selecting trainees. Both concepts revolve around selecting items from a set, but how they handle the order is their key distinction.

  • Permutations concern themselves with arrangements where order matters. For permutations, ABC is different from CAB.
  • Combinations ignore the order, treating ABC and CAB as identical.
In our trainee selection problem, we use combinations because the sequence in which we select the trainees doesn’t change the outcome. Order doesn't affect our choice here, just the selection itself.

Knowing when to apply permutations or combinations is crucial:
  • Use permutations for arranged setups where the sequence is important, like seating arrangements.
  • Use combinations when interested only in selecting items, like forming committees.
Grasping this distinction can simplify problem-solving and eliminate confusion over which formulas to apply.

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Most popular questions from this chapter

On a single toss of a fair coin, the probability of heads is 0.5 and the probability of tails is \(0.5 .\) If you toss a coin twice and get heads on the first toss, are you guaranteed to get tails on the second toss? Explain.

Consider a family with 3 children. Assume the probability that one child is a boy is 0.5 and the probability that one child is a girl is also \(0.5,\) and that the events "boy" and "girl" are independent. (a) List the equally likely events for the gender of the 3 children, from oldest to youngest. (b) What is the probability that all 3 children are male? Notice that the complement of the event "all three children are male" is "at least one of the children is female." Use this information to compute the probability that at least one child is female.

(a) Draw a tree diagram to display all the possible outcomes that can occur when you flip a coin and then toss a die. (b) How many outcomes contain a head and a number greater than \(4 ?\) (c) Probability Extension Assuming the outcomes displayed in the tree diagram are all equally likely, what is the probability that you will get a head and a number greater than 4 when you flip a coin and toss a die?

(a) Explain why -0.41 cannot be the probability of some event. (b) Explain why 1.21 cannot be the probability of some event. (c) Explain why \(120 \%\) cannot be the probability of some event. (d) Can the number 0.56 be the probability of an event? Explain.

A sample space consists of 4 simple events: \(A, B, C, D\) Which events comprise the complement of \(A ?\) Can the sample space be viewed as having two events, \(A\) and \(A^{\text {c }}\) ? Explain.

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