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Chapter 34: Problem 1084

A man and his wife decide to entertain 24 friends by giving 4 dinners with 6 guests each. In how many ways can the first group be chosen?

Short Answer

Expert verified
The first group of 6 guests can be chosen from the 24 friends in 89,297 ways.

Step by step solution

01

Identify the Problem

In this problem, we are asked to find the number of ways to choose the first group of 6 guests out of 24 friends.
02

Apply the Combination Formula

To do this, we need to use the combination formula, which is expressed as: C(n, k) = \( \frac{n!}{k!(n-k)!} \) where C(n, k) represents the number of combinations, n is the total number of elements in the set (24 friends in this case), k is the number of elements to be chosen (6 guests in this case), and ! denotes a factorial.
03

Compute Factorials

Before computing the combination, first, find the factorial values: - For n! (n factorial): \( 24! = 24 脳 23 脳 22 脳 ... 脳 2 脳 1 \) - For \(k!\) (k factorial): \( 6! = 6 脳 5 脳 4 脳 3 脳 2 脳 1 \) - For \((n-k)!\): \( (24-6)! = 18! = 18 脳 17 脳 16 脳 ... 脳 2 脳 1 \)
04

Calculate the Combination

Now, substitute these values into the combination formula: C(24, 6) = \( \frac{24!}{6!(24 - 6)!} \) = \( \frac{24!}{6! 脳 18!} \) To simplify the calculation, cancel out some of the factors in the numerators and denominators: C(24, 6) = \( \frac{24 脳 23 脳 22 脳 21 脳 20 脳 19}{6 脳 5 脳 4 脳 3 脳 2 脳 1} \) Now, do the simplification by dividing: C(24, 6) = \( \frac{(24) 脳 23 脳 22 脳 21 脳 20 脳 19}{(6) 脳 (5) 脳 (4 脳 3 脳 2 脳 1)} \) = \( \frac{23 脳 22 脳 21 脳 20 脳 19}{5 脳 2} \) Notice, dividing a bit further: C(24, 6) = \( \frac{23 脳 11 脳 21 脳 (10) 脳 19}{5} \) = 23 脳 11 脳 21 脳 2 脳 19 Finally, multiply: C(24, 6) = 23 脳 11 脳 21 脳 2 脳 19 = 89,297
05

Interpret the Result

So, there are 89,297 ways to choose the first group of 6 guests out of the man's 24 friends for the dinner.

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