Calculate ways to form a committee of 3 from 8, without division?

[12john]

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: [web link redacted by the Mentors]

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)?

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

 

[andrewkirk]

Your latex doesn't render because on physicsforums you have to use ## as delimiter for in-line maths rather than the usual latex delimiter $.

The reason your answer of 8 x 7 x 6 =336 is wrong is that that is the answer for picking offices, not a committee, ie where order does matter. For a committee, order doesn't matter so you need to divide 336 by the number of different possible orders for three positions, which is 3 x 2 = 6, giving an answer of 56.
That is explained in detail in the text you posted - see the box marked "Important".

But you are not allowed to divide, so you have to think of a different method. A hint lies in an extension of the second solution approach listed under (b) in Example 1.4.13 that you posted. Start by labelling the people as 1 to 8 and then think of different situations where the committee member with lowest number label is 1, 2, 3 etc.
Here's a further hint:
1 : 6+5+4+3+2+1
2 : 5+4+3+2+1
3 : 4+3+2+1
4 : 3+2+1
5 : 2+1
6 : 1
Note that all the numbers to the right of the colons add to 56.

 

[WWGD]

I think wrapping around $$, i.e., double $'s may help it render.

 

[SammyS]

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: [web link redacted by the Mentors]

Orange underline

1. Unquestionably, ${4 \times 3/2} = 3!$ 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, ${{\dfrac{8 \times 7 \times 6}{3!}}}$ = 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?
View attachment 296035

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

Above, find this thread's Post#1

It is essentially a copy from the "page" pointed to by the link given in Post#1:
[web link redacted by the Mentors]

The only material added by OP is the line regarding LaTeX:
" I don't know why my Latex isn't rendering here? Please see "

By the way, I have fixed the LaTeX in the above replied to text.

@12john ,
What is your question in specific?

You merely posted someone else's question posted of some other math site.
And you have shown Zero work of your own.