- #1

scherz0

- 10

- 2

**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?

**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?**

*Here's my surmisal.*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)?

David Patrick,

*Introduction to Counting & Probability*(2005), pp 66-7.