Counting Problem (ways of choosing 3 with conditions)

In summary, the question asks for the number of ways to select 3 light bulbs from a group of 13, consisting of 9 good bulbs and 4 defective bulbs, such that exactly 1 of the selected bulbs is defective. This can be solved using the combination formula C(n,r). The total number of ways to select 3 bulbs is 286, and the number of ways to select only good bulbs is 84. Subtracting these gives 202 ways to get 1 or more defective bulb. To further differentiate between cases, we can use the rule of counting. By selecting 1 defective bulb and 2 good bulbs, we have C(4,1) * C(9,2) = 36 ways
  • #1
mishima
576
43

Homework Statement


Given 9 good light bulbs and 4 defective light bulbs, how many ways can you select 3 such that you get exactly 1 defective bulb?

Homework Equations


C(n,r)

The Attempt at a Solution


I understand total ways to select is C(13,3)=286, and that total ways to select only good bulbs is C(9,3)=84. Subtracting these would give 202 ways to get 1 or more defective bulb. Not sure how to further differentiate these cases...answer provided is 144.

If I just use rule of counting, I can pick from 4 bad, 9 good, then 8 bad in any order which is 288...double the provided answer.
 
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  • #2
mishima said:

Homework Statement


Given 9 good light bulbs and 4 defective light bulbs, how many ways can you select 3 such that you get exactly 1 defective bulb?

Homework Equations


C(n,r)

The Attempt at a Solution


I understand total ways to select is C(13,3)=286, and that total ways to select only good bulbs is C(9,3)=84. Subtracting these would give 202 ways to get 1 or more defective bulb. Not sure how to further differentiate these cases...answer provided is 144.

If I just use rule of counting, I can pick from 4 bad, 9 good, then 8 bad in any order which is 288...double the provided answer.
In how many ways can you select 1 defective bulb?
In how many ways can you select 2 good bulbs?
 
  • #3
C(4,1)*C(9,2). Thank you.
 

FAQ: Counting Problem (ways of choosing 3 with conditions)

1. How do you approach a counting problem that involves choosing 3 items with certain conditions?

To solve a counting problem involving choosing 3 items with conditions, you first need to identify the conditions and determine if they are mutually exclusive or independent. Then, you can use the fundamental counting principle or combinations formula to calculate the total number of ways to choose 3 items that satisfy the given conditions.

2. Can you give an example of a counting problem that involves choosing 3 items with conditions?

One example of a counting problem involving choosing 3 items with conditions is: In how many ways can a committee of 3 students be formed from a group of 10 students, if at least one male and one female student must be included in the committee?

3. What is the fundamental counting principle?

The fundamental counting principle states that if there are n ways to perform the first task and m ways to perform the second task, then there are n x m ways to perform both tasks together. This principle can be extended to more than two tasks as well.

4. When should I use the combinations formula to solve a counting problem?

The combinations formula should be used when the order of the items does not matter. This means that choosing the items in a different order would still result in the same outcome. In other words, the items are being selected as a group rather than individually.

5. What is the difference between permutations and combinations?

Permutations and combinations both involve counting the number of ways to choose items from a group. However, permutations take into account the ordering of the items, while combinations do not. This means that the same items can be chosen in different orders in permutations, but not in combinations.

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