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Magazine Chapter 5: Problem 33
Solve the following: a. \(40 ! / 35 !\) b. \({ }_{7} P_{4}\) c. \({ }_{5} C_{2}\)
Short Answer
Expert verified
a. 78960960, b. 840, c. 10
Step by step solution
01
Analyze part a
Given: \( \frac{40!}{35!} \). We know that the factorial, \(!\), of a number \(n\) is defined as \( n! = n \times (n - 1) \times (n - 2) \times \, \ldots \, \times 1 \). We can simplify the expression by cancelling out the common terms in the numerator and denominator.
02
Simplify part a
The expression \( \frac{40!}{35!} \) simplifies to the product of the integers from 36 to 40:\[ 40 \times 39 \times 38 \times 37 \times 36 \]Calculate that product.
03
Calculate the result for part a
Multiply the numbers:\[ 40 \times 39 = 1560 \]\[ 1560 \times 38 = 59280 \]\[ 59280 \times 37 = 2193360 \]\[ 2193360 \times 36 = 78960960 \]So, \( \frac{40!}{35!} = 78960960 \).
04
Analyze part b
Given: \({ }_{7} P_{4}\). This permutation is the number of ways to arrange 4 items from 7, and is calculated as:\[ {}_7 P_4 = \frac{7!}{(7-4)!} = \frac{7!}{3!} \]
05
Simplify and compute part b
First, calculate:\[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \]Then, calculate:\[ 3! = 3 \times 2 \times 1 = 6 \]Divide:\[ \frac{5040}{6} = 840 \]Thus, \({ }_7 P_4 = 840 \).
06
Analyze part c
Given: \({ }_5 C_2\). This combination is the number of ways to choose 2 items from 5, calculated as:\[ {}_5 C_2 = \frac{5!}{2! \times (5-2)!} = \frac{5!}{2! \times 3!} \]
07
Simplify and compute part c
First, calculate:\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]Then, calculate:\[ 2! = 2 \times 1 = 2 \] and \[ 3! = 3 \times 2 \times 1 = 6 \]Divide:\[ \frac{120}{2 \times 6} = \frac{120}{12} = 10 \]Thus, \({ }_5 C_2 = 10 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factorial
In mathematics, a factorial is a function applied to a positive integer. It is denoted by the symbol !. The factorial of a number n, written as n!, is the product of all positive integers less than or equal to n.
You will often encounter expressions like \( \frac{40!}{35!} \), which simplifies to the product of numbers from 36 to 40. This is because the numbers from 1 to 35 cancel out in both the numerator and the denominator. Understanding how factorial simplifies problems in combinatorics can save time and reduce errors in calculations.
- For example, for n = 5, the calculation is: 5! = 5 脳 4 脳 3 脳 2 脳 1 = 120.
You will often encounter expressions like \( \frac{40!}{35!} \), which simplifies to the product of numbers from 36 to 40. This is because the numbers from 1 to 35 cancel out in both the numerator and the denominator. Understanding how factorial simplifies problems in combinatorics can save time and reduce errors in calculations.
Permutation
A permutation refers to the arrangement of objects in a particular order. When dealing with permutations, order matters. The number of permutations of selecting r items from n distinct items is represented as _{n}P_{r}, and is calculated by the formula:\[{}_{n}P_{r} = \frac{n!}{(n-r)!}\]
This method efficiently calculates the possible arrangements and is widely used in scenarios where order or sequence is significant, such as in organizing races or scheduling.
- For example, \({}_{7}P_{4}\)represents the number of ways to arrange 4 items out of 7, which turns out to be 840 ways.
This method efficiently calculates the possible arrangements and is widely used in scenarios where order or sequence is significant, such as in organizing races or scheduling.
Combination
A combination refers to the selection of items from a larger pool where the order in which they are selected does not matter. Combinations are useful when the sequence of selection is irrelevant, often appearing in problems involving grouping or subsets. The number of combinations for choosing r elements from n is given by:\[{}_{n}C_{r} = \frac{n!}{r!(n-r)!}\]
Combinations are crucial for tasks that involve selecting groups, such as forming a committee or choosing lottery numbers.
- Take \({}_{5}C_{2}\)as an example, where 2 items are chosen from 5, resulting in 10 possible combinations.
Combinations are crucial for tasks that involve selecting groups, such as forming a committee or choosing lottery numbers.
