Math Problem Statement

A committee consists of a president, a treasurer and a secretary. The three posts are to be chosen from a club consisting of 10 people (A, B, C, D, E, F, G, H, I and J). Find the number of ways to form the committee if E and F will not serve together.

Solution

The problem is asking us to find the number of ways to form a committee consisting of a president, a treasurer, and a secretary chosen from 10 people, with the condition that E and F will not serve together.

Step 1: Calculate the total number of ways to form the committee without any restrictions.

Since there are 3 positions (president, treasurer, and secretary), the number of ways to choose these 3 positions from 10 people is the number of permutations of 3 people from 10:

P(10,3)=10!(103)!=10×9×8=720P(10, 3) = \frac{10!}{(10-3)!} = 10 \times 9 \times 8 = 720

Step 2: Calculate the number of ways in which both E and F are in the committee.

If E and F are both in the committee, we must choose 1 more person from the remaining 8 people (A, B, C, D, G, H, I, J). There are 8 ways to choose the third person.

After selecting E, F, and the third person, we can assign them to the 3 positions in the committee. The number of ways to assign these 3 people to the 3 positions is 3!=63! = 6.

Thus, the number of ways in which both E and F serve together is:

8×6=488 \times 6 = 48

Step 3: Subtract the cases where E and F are both in the committee.

To account for the restriction that E and F will not serve together, we subtract the 48 cases where both E and F are serving together from the total number of ways to form the committee:

72048=672720 - 48 = 672

Final Answer:

The number of ways to form the committee if E and F will not serve together is 672.


Would you like to go over any part in more detail, or ask any specific questions?

Here are some related questions you might find helpful:

  1. How do we calculate the number of combinations instead of permutations?
  2. How do we solve similar problems with more restrictions (e.g., more pairs that can't serve together)?
  3. What is the difference between permutations and combinations?
  4. What if instead of 10 people, there were 15? How would the solution change?
  5. How can we solve the problem if we add new positions to the committee?

Tip: When you encounter a problem with restrictions, try solving the problem without restrictions first, then subtract the restricted cases.

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Math Problem Analysis

Mathematical Concepts

Permutations
Combinatorics
Restrictions

Formulas

Permutation formula: P(n, r) = n! / (n - r)!
Factorial: n! = n × (n-1) × ... × 1

Theorems

Fundamental Counting Principle

Suitable Grade Level

Grades 9-12