Combination rule: if 4 people are selected from group....

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stacey

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May 6, 2007
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Hello,
I am taking an online elementary statistics.
I am trying to understand counting numbers.

if 4 people are selected from group 10 you use the combination rule

10C4=10!/(10-4) !4! = 10!/6!4!

i don't understand how they get the answer from 10! /6!*4! =210

What am I missing with the factorial and understanding this formula?

Stacey
 
Here is another Counting numbers rule?

How many 3 digit ID tags can be made if digits can be used more than once. If the first digit must be 5 and repetitions are not permitted.
since there are three spaces to fill and 10 choices for each space: the solution is
10*10*10=1000
How many different 3-digit ID tags can be made if the first digit must be 5 and repetitions are not permitted?
since there are three spaces to fill, the first space must be a five and there is one less choice for eachsubsequent space. the solution is: 1*9*8=72

1. What is "the first digit must be 5" how does this work in?
2. can someone explain this in basic terms?
Stacey
 
Re: Combination rule

stacey said:
10C4=10!/(10-4) !4! = 10!/6!4!
i don't understand how they get the answer from 10! /6!*4! =210
Click to expand...
First of all, you need parentheses: 10! /(6!*4!)

10! = 3628800

6! * 4! = 720 * 24 = 17280

Now simply divide.
 
stacey said:
How many different 3-digit ID tags can be made if the first digit must be 5 and repetitions are not permitted?
since there are three spaces to fill, the first space must be a five and there is one less choice for each subsequent space. the solution is: 1*9*8=72

1. What is "the first digit must be 5" how does this work in?
2. can someone explain this in basic terms?
Stacey
Click to expand...
Your question is quite unclear: what do you want to know?
If the 1st digit must be 5, then the other digits cannot be 5: no repetitions.

Then there are 9 possible digits as 2nd digit: all digits except the 5.

When a digit is chosen as 2nd digit, then there are 8 possible digits
as 3rd digit: all digits except the 5 and the one chosen for 2nd.

Is that what you were asking?

AND: why are you using the expression "counting numbers"?
 
Re: Combination rule

Hello, Stacey!

With an online course, you probably didn't get much practice with factorials.

\(\displaystyle _{_{10}}C_{_4}\:=\:\frac{10!}{6!\,4!} \:=\:210\)

i don't understand how they get the answer from \(\displaystyle \frac{10!}{6!\,4!}\)
Click to expand...

After a few problems, you'll get faster at this . . .

We know what factorials are, so we can write out that fraction.

. . \(\displaystyle \L\frac{10!}{6!\,4!} \:=\:\frac{10\cdot9\cdot\8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{(6\cdot5\cdot4\cdot3\cdot3\cdot2\cdot1)(4\cdot3\cdot2\cdot1)}\)

Look at what cancels immediately: \(\displaystyle \L\,\frac{10!}{6!\,4!} \:=\:\frac{10\cdot9\cdot\8\cdot7\cdot\not6\cdot\not5\cdot\not4\cdot\not3\cdot\not2\cdot\not1}{(\not6\cdot\not5\cdot\not4\cdot\not3\cdot\not2\cdot\not1)(4\cdot3\cdot2\cdot1)}\)

. . (If we anticipate this cancelling, we can save a lot of writing!) **


Then we reduce "what's left": \(\displaystyle \L\:\frac{10\cdot\not9^{^3}\cdot\not8\cdot7}{\not4\cdot\not3\cdot\not2\cdot1} \:=\:210\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

** . Consider the value of: \(\displaystyle \L\:\frac{50!}{48!}\)

We know the numerator is: \(\displaystyle \,50\cdot49\cdot48\cdots3\cdot2\cdot1\)
. - and the denominator is: \(\displaystyle \,48\cdot47\cdot46\cdots3\cdot2\cdot1\)

But we are not going to write all those factors . . . nor multiply them.

We can see (can't we?) that the \(\displaystyle 48!\) in the denominator
. . cancels out most of the numerator.

We say to ourselves: "We have 50 times 49,
. . and 48-on-down is cancelled by the 48! in the denominator."

And there is the value: \(\displaystyle \L\:50\cdot49 \:=\:2450\)

 
Denis,
Your correct on what they provide and don't provide for online. I enjoy the course just working hard to understand it. I figured out the Factorial this morning before work :) but your reply is very helpful. I posted last night on the discussion board at school but no response. No one uses it. I thank you for your time. As for the counting numbers post. That came word for word of f the video. 3 digit ID tags can be made? okay three digits 123, 333, 234, if the first digit must be 5? I will let you know if I get a reply.

Stacey
 
How many 3 digit ID tags can be made if digits can be used more than once-if the first digit must be 5 and repetitions are not permitted?

10*10*10=1000
part 2 since there are three spaces to fill the first space must be 5 and there is one less choice for each subsquent space?
answer 1*9*8=72
The five is throwing me for a loop?
Stacey
 
stacey said:
part 2 since there are three spaces to fill the first space must be 5 and there is one less choice for each subsquent space?
answer 1*9*8=72
The five is throwing me for a loop?
Stacey
Click to expand...
WHY is it throwing you for a loop?
You seem to be complicating something simple:
if the 1st digit of a 3 digit number is a 5, then the numbers go from 500 to 599:
that's it.
 
My Mother was born in Ottawa. I am half Canadian
I never did well in math and decided to try and better my self.
So you saying the first digit is 5_ _ -5 _ _
I will have to thin about that one
thanks again
night night
 
stacey said:
My Mother was born in Ottawa. I am half Canadian
I never did well in math and decided to try and better my self.
So you saying the first digit is 5_ _ -5 _ _
I will have to thin about that one
Click to expand...
...so half of you is LUCKY :roll:
Ask Mom if she remembers the Sparks St Mall; or Mooney's Bay.

Yes, that's what I'm saying; SO is your problem:
> How many different 3-digit ID tags can be made if the first digit must
> be 5 and repetitions are not permitted?

Quite clear, ain't it: 1st digit must be 5, next 2 cannot be 5 (no repetitions)
and cannot be the same (no repetitions); first few:
500 : no, two 0's
501 : yes
502 : yes
503 : yes
504 : yes
505 : no, two 5's
506 to 510 : yes
511 : no
and so on....
Kapish kapush?
 
:) All of me is good.
I got it.
My Mother spoke of Canada foundly.
I have cousins there and visit often.
I would ask her but she passed away sevreal years now.
thanks for your help my post on the Discussion board has still not been answered. Quiz 4 tomorrow.
Stacey
 
For on line courses you do all your assignments through "Blackboard" there you find your assignments, quizes, exams and a discussion board to of course discuss math problems. So you have been my DB the last few days.
 
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