Question

In Maricopa County, 5 persons are to be elected to the Board of Supervisors. If 8 persons are candidates, how many different arrangements are possible.


6,720

336

56
2.From a committee consisting of 8 men and 5 women, a subcommittee is formed consisting of 4 men and 3 women. How many different subcommittees are possible?
80

700

1,716
How many straight lines can be drawn through 20 points, of which no three lie on any one straight line?
190

380

1,140

Answer 1

@mathmate Please help quickly!

Answer 2

In Maricopa County,
8 choose 5, or 8C5 assuming all posts are identical.

sub-committee
Choose 4 men out of 8, and 3 women out of 5.
Multiply the results together for the total number of selections.
8 choose 4 x 5 choose 3

straight lines
Takes two points to draw a straight line, and order is not important, so 20 choose 2, or 20C2

Answer 3

What does that mean "8 choose 4 x 5 choose 3" @mathmate

Answer 4

It means the product (8C4)(5C3)
I call 8C4 as 8 choose 4 making it easier to remember.


where \[ nCr = \frac{n!}{(n-r)!r!} \]

Answer 5

Oh okay. So what would my equation look like?

Answer 6

@mathmate

Answer 7

which equation?

Answer 8

Idk I'm so confused! :/

Answer 9

@mathmate

Answer 10

Which of the three problems are you working on?

Answer 11

Let's start with the first one @mathmate

Answer 12

When order is not important, such as when we choose 5 people out of 8 for a committee (all posts identical, no hierarchy), then we use combination,
\[8C5 = \frac{8!}{(8-5)!5!}\]
to denote the number of ways the choices can be made.
The reasoning goes as follows:
There are 8 choices for the first job, 7 choices for the second, and 6 for the third and so on until 4 for the fifth job.
that gives 8*7*6*5*4 ways.
But these 5 people could have been chosen in a different order in any of the 5! ways, so the total number of ways must be reduced by this factor, so
8*7*6*5*4/(1*2*3*4*5)
=8C5.

Answer 13

so 56?

Answer 14

@mathmate okay will you help me with the second one now?

Answer 15

It's like the first, except you choose first 4 men from 8, then 3 women from 5.
After that, you would multiply them together, because they would make different subcommittees.

Answer 16

It's confusing :(

Answer 17

@mathmate

Answer 18

For number two, there are two parts.
Part 1 is for men (choose 4 men from 8), and part 2 is for women (choose 3 women from 5).
Can you do the two parts separately, using 8C4 and 5C3.