How to intuit, construe multiplicands and multiplicators $\le 10$ resulting from ${\displaystyle (\genfrac{}{}{0ex}{}{p}{c})}$, WITHOUT division or factorials? [duplicate]

I grok, am NOT asking about, the answers below. Rather — how can I deduce and intuit the multiplicands and multiplicators $\le 10$, resulting from simplifying ${\displaystyle (\genfrac{}{}{0ex}{}{p\text{ people}}{c\text{-person committee}})}$ DIRECTLY? WITHOUT division or factorials!

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1. Unquestionably, $4\times 3/2=3!$. But how can I construe, intuit $3!$ DIRECTLY? Note that $3!$ wasn't one of the original numbers (2, 4). 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. Did I surmise correctly?

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2. Unquestionably, ${\displaystyle \frac{8\times 7\times 6}{3!}}=8\times 7$. But how can I construe, intuit $8\times 7$ DIRECTLY? Note that $7$ wasn't one of the original numbers of 3 or 8. What does $7$ mean?

Here's my surmisal. You can pick the 1st committee member in $8$ ways, and the 2nd member in $7$ ways. Then can't you pick the 3rd member in 6 ways? By this Constructive Counting (David Patrick, p 38 bottom), the answer ought be 8 × 7 × 6? Why didn't 6 appear as a multiplicator?

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)?

David Patrick, BS Math & Computer Science, MS Math (Carnegie Mellon), PhD Math (MIT). Introduction to Counting & Probability 2005, pp 66-7.

Examples 3 and 4

3. I know the ways to form a committee of 2 from 8 $={\displaystyle (\genfrac{}{}{0ex}{}{8}{2})}={\displaystyle \frac{8\cdot 7}{2!}}=4\times 7.$ But how can I construe, intuit $4\times 7$ DIRECTLY? Note that $4,7$ aren't the original numbers of 2, 8.

4. I know the ways to form a committee of 3 from 10 $={\displaystyle (\genfrac{}{}{0ex}{}{10}{3})}={\displaystyle \frac{10\cdot 9\cdot 8}{3!}}=10\times 3\times 4.$ But how can I construe, intuit $10\times 3\times 4$ DIRECTLY? I'm surmising that $10$ refers to the original 10, and $3$ to the chosen 3. But what does $4$ mean?

posted about 3 years ago

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3y ago

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