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Chapter 6: Problem 18

How many ordered sequences are possible that contain three objects chosen from seven?

Short Answer

Expert verified
There are 210 possible ordered sequences containing three objects chosen from seven, as determined by the permutation formula without repetition: \( 7P3 = \frac{7!}{(7-3)!} = \frac{7!}{4!} = \frac{5040}{24} = 210 \).

Step by step solution

01

Understand the problem

You are asked to find the number of ordered sequences with three different objects chosen from seven. In this case, the order of the objects matters, and you can't choose an object more than once. This is a permutation problem without repetition.
02

Recall the formula for permutations without repetition

The formula for finding the number of permutations without repetition is given by: nPr = n! / (n-r)! Where n is the total number of objects, r is the number of objects you want to choose, and n! means the factorial of n, which is the product of all positive integers up to n.
03

Apply the formula to the given problem

In this problem, you have 7 objects (n = 7) and you want to choose 3 (r = 3). So, you need to find the number of permutations without repetition by applying the formula: \( 7P3 = \frac{7!}{(7-3)!} \)
04

Calculate the factorials

First, calculate the factorials: 7! = 7 脳 6 脳 5 脳 4 脳 3 脳 2 脳 1 = 5040 (7-3)! = 4 ! = 4 脳 3 脳 2 脳 1 = 24
05

Complete the calculation

Now, finish the calculation by dividing the factorials: \( 7P3 = \frac{7!}{4!} = \frac{5040}{24} = 210 \)
06

Interpret the result

There are 210 possible ordered sequences containing three objects chosen from seven.

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