/*! 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 34 Solve the following: a. \(20 !... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the following: a. \(20 ! / 17 !\) b. \({ }_{9} P_{3}\) c. \({ }_{7} C_{2}\)

Short Answer

Expert verified
a. 6840; b. 504; c. 21.

Step by step solution

01

Simplify Factorials

To simplify \( \frac{20!}{17!} \), note that the factorial notation implies that \( 20! = 20 \times 19 \times 18 \times 17! \). Thus, the \( 17! \) in the numerator and denominator cancels out, leaving us with \( 20 \times 19 \times 18 \). Calculate the product: \[ 20 \times 19 = 380 \] \[ 380 \times 18 = 6840 \] Therefore, \( \frac{20!}{17!} = 6840 \).
02

Calculate Permutation

The permutation \( {}_9 P_3 \) is calculated using the formula \( nP_k = \frac{n!}{(n-k)!} \). Here, \( n = 9 \) and \( k = 3 \), so:\[ {}_9 P_3 = \frac{9!}{(9-3)!} = \frac{9!}{6!} \]Simplifying: \[ 9! = 9 \times 8 \times 7 \times 6! \] then \( 6! \) cancels out, leaving \[ 9 \times 8 \times 7 = 504 \] .So, \( {}_9 P_3 = 504 \).
03

Calculate Combination

The combination \( {}_7 C_2 \) is calculated using the formula \( nC_k = \frac{n!}{k!(n-k)!} \). Here, \( n = 7 \) and \( k = 2 \), so:\[ {}_7 C_2 = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} \]Simplifying, we have \[ 7! = 7 \times 6 \times 5! \] so \[ \frac{7 \times 6 \times 5!}{2!5!} \] which cancels the \( 5! \), leaving:\[ \frac{7 \times 6}{2 \times 1} = \frac{42}{2} = 21 \] So, \( {}_7 C_2 = 21 \).

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Key Concepts

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

Factorials
Factorials are an essential part of combinatorics and statistics, used to calculate permutations and combinations. A factorial, represented by the symbol "!", is the product of an integer and all the integers below it, down to 1. For example, the factorial of 5, written as 5!, is calculated as \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Factorials grow rapidly as the number increases, making them useful in counting problems involving large sets. They are especially helpful in simplifying expressions involving permutations and combinations. For instance, to simplify the expression \( \frac{20!}{17!} \), we recognize that the sequence from 17! to 1 cancels out, leaving us with just \( 20 \times 19 \times 18 \). This technique is widely used to handle higher factorial numbers without extensive calculations.

Key Points About Factorials:
  • Factorial of 0 is defined as 1, i.e., 0! = 1.
  • Factorials are central to permutations and combinations.
  • They simplify to make calculations more manageable, especially when common terms can be canceled.
Permutations
Permutations relate to the arrangement of a set of objects where order does matter. The number of permutations of n objects taken k at a time is given by the formula \( nP_k = \frac{n!}{(n-k)!} \). This formula calculates how many different ways you can arrange k items out of n, considering the sequence to be crucial.

Let's delve into an example using \({ }_9 P_3\). Here, we're calculating the permutations of 9 objects taken 3 at a time. Following our formula, we determine it as \( \frac{9!}{6!} \) which simplifies directly to \( 9 \times 8 \times 7 \) after canceling out 6! from both numerator and denominator. Thus, \({ }_9 P_3 = 504\). This indicates there are 504 different sequences three items can be arranged from nine.

Understanding Permutations:
  • Order of arrangement is crucial in permutations.
  • The number of possibilities reduces as objects are chosen and arranged sequentially.
  • Helpful in scenarios like creating passwords, arranging objects, or seating people.
Combinations
While permutations consider the order vital, combinations count arrangements without regard to the sequence. To calculate combinations, we use the formula \( nC_k = \frac{n!}{k!(n-k)!} \). This formula finds the number of ways to choose k objects from a set of n, where order is irrelevant.

Consider the combination \({ }_7 C_2\). We are finding the number of ways to choose 2 items from 7 without considering the order. Using our formula, it simplifies to \( \frac{7!}{2!5!} \). Upon performing the cancellations in the factorials, we're left with \( \frac{7 \times 6}{2 \times 1} = 21 \). Hence, there are 21 different ways to pick 2 objects from 7.

Insights on Combinations:
  • Order does not affect the count in combinations.
  • Useful in selecting groups from a larger set, such as forming teams, choosing lottery numbers.
  • Typically result in fewer outcomes than permutations due to order neglect.

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