How to intuit, construe multiplicands and multiplicators $\le 10$ resulting from $\dbinom pc$, WITHOUT division or factorials? [duplicate]
Closed as duplicate by Peter Taylor on Jan 25, 2022 at 09:20
This question has been addressed elsewhere. See: How can I deduce which operation removes redundacies?
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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 $\dbinom {p \text{ people}}{c\text{person committee}}$ DIRECTLY? WITHOUT division or factorials!
Orange underline
 Unquestionably, $\color{darkorange}{4 \times 3/2 = 3!}$. But how can I construe, intuit $\color{darkorange}{3!}$ DIRECTLY? Note that $\color{darkorange}{3!}$ wasn't one of the original numbers (2, 4). What does $\color{darkorange}{3!}$ 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 2person committee in 3 ways. Did I surmise correctly?
Red underline
 Unquestionably, $\color{red}{\dfrac{8 \times 7 \times 6}{3!} = 8 × 7}$. But how can I construe, intuit $\color{red}{8 × 7}$ DIRECTLY? Note that $\color{red}{7 } $ wasn't one of the original numbers of 3 or 8. What does $\color{red}{7 } $ mean?
Here's my surmisal. You can pick the 1st committee member in $\color{red}{8} $ ways, and the 2nd member in $\color{red}{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 2person 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 667.
Examples 3 and 4

I know the ways to form a committee of 2 from 8 $= \dbinom8 2 = \dfrac{8\cdot7}{2!} = \color{violet}{4 × 7}.$ But how can I construe, intuit $\color{violet}{4 × 7}$ DIRECTLY? Note that $\color{violet}{4, 7}$ aren't the original numbers of 2, 8.

I know the ways to form a committee of 3 from 10 $= \dbinom {10} 3 = \dfrac{10\cdot9\cdot8}{3!} = \color{springgreen}{10 × 3 × 4}.$ But how can I construe, intuit $\color{springgreen}{10 × 3 × 4}$ DIRECTLY? I'm surmising that $\color{springgreen}{10}$ refers to the original 10, and $\color{springgreen}{3}$ to the chosen 3. But what does $\color{springgreen}{4}$ mean?
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