Problem 33 - Basic Statistics for Business and Economics with Student

Question

Solve the following: a. \(40 ! / 35 !\) b. \({ }_{7} P_{4}\) c. \({ }_{5} C_{2}\)

Step-by-Step Solution

Answer

a. 7893600, b. 840, c. 10.

Step 1: Understanding Factorials

The factorial of a number, denoted by \( n! \), is the product of all positive integers less than or equal to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).

Step 2: Evaluating Factorial Division

To evaluate \( \frac{40!}{35!} \), note that \( 40! = 40 \times 39 \times 38 \times 37 \times 36 \times 35! \). Hence, the \( 35! \) terms in the numerator and denominator cancel out, leaving \( 40 \times 39 \times 38 \times 37 \times 36 \).

Step 3: Calculation of Remaining Product

Multiply the remaining terms: \( 40 \times 39 \times 38 \times 37 \times 36 = 7893600 \). Thus, \( \frac{40!}{35!} = 7893600 \).

Step 4: Understanding Permutations

\( {}_n P_r \) represents permutations of \( n \) objects taken \( r \) at a time and is calculated as \( \frac{n!}{(n-r)!} \).

Step 5: Evaluating Permutation

For \( {}_7 P_4 \), apply the formula: \( \frac{7!}{(7-4)!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 840 \).

Step 6: Understanding Combinations

\( {}_n C_r \) represents combinations of \( n \) objects taken \( r \) at a time and is calculated with the formula \( \frac{n!}{r!(n-r)!} \).

Step 7: Evaluating Combination

For \( {}_5 C_2 \), apply the formula: \( \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10 \).

Key Concepts

Permutations

Permutations are all about arranging objects in a specific sequence or order. This concept is crucial when you need to determine the number of ways to arrange a set of items. For calculating permutations, you use the formula \[ {}_n P_r = \frac{n!}{(n-r)!} \]where

• \( n \) is the total number of objects.
• \( r \) is the number of objects we want to arrange.

In simpler terms, permutations answer the question: "In how many different ways can I arrange \( r \) items from a group of \( n \)?"Let's consider the example of finding \( {}_7 P_4 \), which represents the number of ways to arrange 4 objects out of 7 distinct objects. Using the formula:\[{}_7 P_4 = \frac{7!}{(7-4)!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 840\]This calculation shows there are 840 possible permutations of 4 objects chosen from 7. Permutations emphasize order, and every different sequence counts as a distinct permutation.

Combinations

Combinations focus on selecting items without considering the order. They are particularly useful when the sequence of chosen objects doesn't matter, just the selection itself. The formula for combinations is given by:\[{}_n C_r = \frac{n!}{r!(n-r)!}\]where

• \( n \) is the total number of objects available.
• \( r \) is the number of objects to be selected.

The calculation provides the number of ways to choose \( r \) objects from \( n \) objects without regard to order.Let's break down the calculation of \( {}_5 C_2 \). The aim is to find how many ways we can select 2 items from a set of 5. Plugging the values into the formula gives:\[{}_5 C_2 = \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10\]Thus, there are 10 different combinations possible. Unlike permutations, combinations do not consider different sequences as unique. It's simply about which items are chosen.

Mathematical Operations

Mathematical operations form the backbone of solving problems involving permutations and combinations. These operations include addition, subtraction, multiplication, and division, which are processes that enable us to calculate factorial values and simplify expressions.Factorials (\( n! \)) are fundamental in these operations because they represent the product of integers from 1 through \( n \). For instance, when faced with an expression like \( \frac{40!}{35!} \),the idea is to simplify it by recognizing that \( 40! = 40 \times 39 \times 38 \times 37 \times 36 \times 35! \).Cancelling out the \( 35! \)in numerator and denominator reduces the expression to \( 40 \times 39 \times 38 \times 37 \times 36 \),making it a simple multiplication problem:\[40 \times 39 \times 38 \times 37 \times 36 = 7893600\]The solution to such factorial division problems often requires understanding the basic multiplication operation and how terms can cancel each other out.By mastering these operations, tackling more complex algebraic expressions becomes easier, forming a solid mathematical foundation for various applications.