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

[scherz0]

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?

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

[Genady]

It is just a coincidence that the 6 in numerator and the 6 in denominator cancel, in this case. Take 3 out of 9 instead of out of 8. You get 9*8*7/6. You got to divide.

[OldChemE]

overly complicated.  To make a committee of 3 from 8:  You have 8 choices for member #1, then 7 choices for the next member, then 6 for the third member.  8 x 7 x 6

[Genady]

13 minutes ago, OldChemE said:

overly complicated.  To make a committee of 3 from 8:  You have 8 choices for member #1, then 7 choices for the next member, then 6 for the third member.  8 x 7 x 6

... and then you need to divide it by 3! to get the correct number of ways.

[joigus]

You've got 8 people. Count the number of ways you can make them sit in 8 chairs. It's 8!

Now, every time you sit them, secretly, 8-3=5 of the chairs you declare irrelevant (not belonging to the committee), so you divide by 5! to discount those irrelevant reshufflings. They're not in the committee anyway.

Now you're down to the number of ways in which you can select groups of 3 people from a group of 8 people and make them sit on 3 different chairs. The committee chairs. You've still overcounted, because it doesn't matter where they sit.

So you divide by 3! and you're down to the combinatory number you need. It doesn't matter where they sit, and it doesn't matter where excluded people sit. It only matters who was selected for the committee.

[joigus]

Sorry. You meant without division. I can't think of a way to do that that's as transparent as dividing by the factorials of whatever is irrelevant.

[Genady]

Yes, you can do without division, by a "Richard Feynman's method": make a list of all possible ways to form groups of 3 from 8 elements, then count them.  

[studiot]

14 hours ago, OldChemE said:

overly complicated.  To make a committee of 3 from 8:  You have 8 choices for member #1, then 7 choices for the next member, then 6 for the third member.  8 x 7 x 6

Posted in error.

 

Is this meant to be a puzzle ?

I think one way is to place all the members in order thus

A B C D E F G H

Then form committees from the first member plus two others.
Since there are 8 members you need (8-1 -1 = 6 ) ways to do this for the first, second and one other member.
Also for the first, third and one other member there are 5 ways.
For the first, fourth and one are there are 4 ways.
For the first, fifth and one other there are 3 ways.
For the first, sixth and one other there are 2 ways.
For the first, seventh and pone other there is 1 way.
A total of 6+5+4+3+2+1 distinct ways

Then discard the first member and start again with a committe from 7 members which will have

5+4+3+2+1 ways distinct from the first set

and so on forming the following table of ways

[math]\begin{array}{*{20}{c}}
   6 \hfill & 5 \hfill & 4 \hfill & 3 \hfill & 2 \hfill & 1 \hfill  \\
   {} \hfill & 5 \hfill & 4 \hfill & 3 \hfill & 2 \hfill & 1 \hfill  \\
   {} \hfill & {} \hfill & 4 \hfill & 3 \hfill & 2 \hfill & 1 \hfill  \\
   {} \hfill & {} \hfill & {} \hfill & 3 \hfill & 2 \hfill & 1 \hfill  \\
   {} \hfill & {} \hfill & {} \hfill & {} \hfill & 2 \hfill & 1 \hfill  \\
   {} \hfill & {} \hfill & {} \hfill & {} \hfill & {} \hfill & 1 \hfill  \\
\end{array}[/math]

Add them all up to obtain the answer.

 

Edited January 31, 2022 by studiot

[joigus]

The thing I don't quite get is, why would you wanna do that, if it's more cumbersome and much, much less transparent?

Dividing by factorials to correct for redundancy is the safest quickest way to master combinatorics. Except those in which repetition is allowed, and are driven by powers.

[studiot]

10 minutes ago, joigus said:

The thing I don't quite get is, why would you wanna do that, if it's more cumbersome and much, much less transparent?

Dividing by factorials to correct for redundancy is the safest quickest way to master combinatorics. Except those in which repetition is allowed, and are driven by powers.

Patterns ?

For m members and n selections it is a double product series of terms, one ascending, one descending

In this case we have

1*6 + 2*5 + 3*4 + 4*3 + 5*2 + 6*1

 6  +    10  +  12  +   12 +   10 +   6

[OldChemE]

3 hours ago, studiot said:

Posted in error.

yep