Calculate ways to form a committee of 3 from 8, DIRECTLY WITHOUT ÷?

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In summary, the conversation discusses the calculation of combinations and permutations without using division. The first part talks about using factorials to calculate the number of ways to choose a committee. The second part discusses the same concept but with a group of 8 people. The problem being discussed is problem 4.1(b) from David Patrick's book "Introduction to Counting & Probability".
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scherz0
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I grok, am NOT asking about, the answers below. Rather, how can I calculate the final answer DIRECTLY, without division? I don't know why my Latex isn't rendering here? Please see https://math.codidact.com/posts/285679.

Orange underline

1. Unquestionably, $\color{#FFA500}{4 \times 3/2} = 3!$ But how can I construe 3! DIRECTLY WITHOUT DIVISION? What does 3! mean?

Here's my surmisal. Blitzstein's solution hints to this calculation, but he didn't write 3! explicitly. You fix the first person. Then of the 3 people left, you can pick the 2nd person in your 2-person committee in 3 ways. Is this correct?

Red underline

2. Unquestionably, ${\color{red}{\dfrac{8 \times 7 \times 6}{3!}}}$ = 8 × 7. But how can I construe 8 × 7 DIRECTLY WITHOUT DIVISION ? What does 8 × 7 mean?

Here's my surmisal. You can pick the 1st committee member in 8 ways, and the 2nd member in 7 ways. But then can't you pick the 3rd member in 6 ways? By this Constructive Counting (p 38 bottom), the answer ought be 8 × 7 × 6? Why isn't there 6?

Problem 4.1:

(b) In how many ways can a 2-person committee be chosen from a group of 4 people (where the order in which we choose the 2 people doesn't matter)?

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David Patrick, Introduction to Counting & Probability (2005), pp 66-7.
 
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I don't understand why you would WANT to avoid dividing! There are 8 ways to choose one person, say "A", then 7 people left to choose from so 7 ways to choose the second person, "B", then 6 ways to choose the third person, "C". There are 8*7*6 ways to choose "ABC" IN THAT ORDER. (There are 8*7*6= 336 ways to choose a "president, vice-president, and treasurer".) But "ABC", "ACB", "BAC", "BCA", "CAB", "CBA" would be the same three people on a "committee".
 

Related to Calculate ways to form a committee of 3 from 8, DIRECTLY WITHOUT ÷?

1. How many ways can a committee of 3 be formed from a group of 8 people?

There are 56 ways to form a committee of 3 from a group of 8 people, without using division.

2. Can you explain the process for calculating the number of ways to form a committee of 3 from 8 people?

To calculate the number of ways to form a committee of 3 from a group of 8 people, we use the combination formula: nCr = n! / (r! * (n-r)!), where n is the total number of people and r is the number of people we want in the committee. In this case, n = 8 and r = 3. Therefore, nCr = 8! / (3! * (8-3)!) = 56.

3. Why can't we use division to calculate the number of ways to form a committee of 3 from 8 people?

Division is not used to calculate the number of ways to form a committee of 3 from 8 people because it does not take into account the order in which the people are chosen. In other words, division does not consider the fact that choosing person A, B, and C is different from choosing person C, B, and A. Therefore, we use the combination formula to account for all possible combinations.

4. How does the number of people in the group affect the number of ways to form a committee of 3?

The number of people in the group directly affects the number of ways to form a committee of 3. As the number of people in the group increases, the number of ways to form a committee of 3 also increases. This is because there are more people to choose from, resulting in more possible combinations.

5. Is there a faster way to calculate the number of ways to form a committee of 3 from 8 people?

Yes, there is a faster way to calculate the number of ways to form a committee of 3 from 8 people. We can use the shortcut formula: nCr = (n * (n-1) * (n-2)) / (r * (r-1) * (r-2)). Plugging in n = 8 and r = 3, we get (8 * 7 * 6) / (3 * 2 * 1) = 56. This formula is useful for larger numbers, as it reduces the number of calculations needed.

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