Math Problem Statement

In how many different ways can the director of research laboratory choose 2 chemists from among 7 applicants and 3 physicists from among 9 applicants? (3 points)

Solution

To find how many different ways the director of a research laboratory can choose 2 chemists from among 7 applicants and 3 physicists from among 9 applicants, we can solve this using the concept of combinations.

The formula for combinations is given by:

C(n,r)=n!r!(nr)!C(n, r) = \frac{n!}{r!(n - r)!}

Where:

  • nn is the total number of applicants,
  • rr is the number of people to choose,
  • C(n,r)C(n, r) represents the number of ways to choose rr applicants from nn.

Step 1: Choosing 2 chemists from 7 applicants

The number of ways to choose 2 chemists from 7 applicants is:

C(7,2)=7!2!(72)!=7×62×1=21C(7, 2) = \frac{7!}{2!(7 - 2)!} = \frac{7 \times 6}{2 \times 1} = 21

Step 2: Choosing 3 physicists from 9 applicants

The number of ways to choose 3 physicists from 9 applicants is:

C(9,3)=9!3!(93)!=9×8×73×2×1=84C(9, 3) = \frac{9!}{3!(9 - 3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84

Step 3: Total number of ways

Since the two events (choosing chemists and choosing physicists) are independent, we multiply the two results to get the total number of ways:

21×84=176421 \times 84 = 1764

Thus, the director can choose 2 chemists and 3 physicists in 1,764 different ways.


Do you want more details or have any other questions?

Here are 5 questions to further expand on this:

  1. How would the result change if the director had to choose 3 chemists instead of 2?
  2. What is the total number of ways to choose at least 1 chemist and 1 physicist?
  3. Can combinations be used for selecting people when order matters?
  4. What is the difference between combinations and permutations?
  5. How does the formula for combinations change if there are identical applicants?

Tip: When dealing with combinations, order does not matter, unlike in permutations.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Combinations

Formulas

C(n, r) = n! / (r!(n - r)!)

Theorems

Combination theorem

Suitable Grade Level

Grades 10-12