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Chapter 6: Problem 19

How many unordered sets are possible that contain three objects chosen from seven?

Short Answer

Expert verified
There are 35 unordered sets possible that contain three objects chosen from seven.

Step by step solution

01

Understand the problem and identify the formula

We are given 7 objects and we have to find the number of unordered sets containing 3 objects. We will use the formula for combinations, denoted as C(n, r), where n is the total number of items and r is the number of items we want to choose: C(n, r) = \( \frac{n!}{(n-r)!r!} \) In our case, n = 7 (total objects) and r = 3 (objects we need to choose).
02

Apply the formula and calculate the combinations

Now plug in n = 7 and r = 3 into the formula: C(7, 3) = \( \frac{7!}{(7-3)!3!} \)
03

Simplify the expression

Before calculating the combinations, let's simplify the expression: C(7, 3) = \( \frac{7!}{(4)!3!} \)
04

Calculate the factorials

Calculate the factorials for the expression: 7! = 7 脳 6 脳 5 脳 4 脳 3 脳 2 脳 1 = 5040 4! = 4 脳 3 脳 2 脳 1 = 24 3! = 3 脳 2 脳 1 = 6 Now replace the factorials in the expression: C(7, 3) = \( \frac{5040}{24 脳 6} \)
05

Compute the result

Finally, calculate the result for the expression: C(7, 3) = \( \frac{5040}{144} \) = 35 So, there are 35 unordered sets possible that contain three objects chosen from seven.

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