factorial/probability problem (defective light bulbs)

[Guest]

I can't figure out how they canceled the factorials in this problem. Here it is:

How many bulbs should be examined so that the probability of finding at least 1 bad bulb exceeds 1/2?

I know I can do this problem an easier way, but I want to learn it this way:

(2)(48)  +  (2)(48)
(1)(n )     (2)(n )
-------------------
        (50)
        (n )

****Just in case that is hard to read, this is what I mean:

C(2, 1) × C(48, n) + C(2, 2) × C(48, n)
---------------------------------------
              C(50, n)

What I get next is too hard to type up in here, but basically I'm wondering how I would expand (48 - n)! I can't find the formula for expanding it. I know that 3! would be 1*2*3, but how would I do (48 - n)! ? Would it be:

. . .(1 - n)(2 - n)(3 - n)...(48 - n)

Thanks
Take care,
Beckie

 

[pka]

Why not just type the actual problem as it is given.
Let us see what it is asking.

 

[stapel]

...how would I do (48 - n)! ? Would it be:

. . .(1 - n)(2 - n)(3 - n)...(48 - n)

The factorial works for this expression just like the one in your other thread did: Start with the given value (in this case, 48 - n) and multiply by all the preceeding whole numbers (in this case, (48 - n) - 1, (48 - n) - 2), and so forth, down to (48 - n) - (48 - n + 1) = 1).

Eliz.

 

[Guest]

I still dont understand.

What is the formula for expanding (3-n)! ??????

Does it look like this:
(1-n)(2-n)(3-n)

I'm really confused about this problem. I really need to see it so I can do all the steps no matter how long it takes. I don't like to take short cuts.

Thanks
Take care,
Beckie

 

[stapel]

I still dont understand. What is the formula for expanding (3-n)! ?

You would expand this factorial in just the same way as you have expanded every other factorial: Start with "3 - n", and count down (that means "subtract 1 from the last number") until you get down to "1". Then multiply them all together.

Does it look like this: (1-n)(2-n)(3-n)

Since 2 - n is one greater than 3 - n, and since you are supposed to count down to 1 (not count up to whatever), then your expression cannot be the expansion of the given factorial.

Eliz.

 

[Guest]

Can someone please give me the answer to (3-n)!

No one is answering me directly. I know you guys are trying to help me but I still dont see it.

It sounds like you guys are saying to do this:

(3-n) - 1... then what do I multiply the next value or add it or subtract it?

(3-n) - 1 + (2- n) - 1 + (1-n) - 1

sorry if I'm coming off so strong, but I have been working on this problem for like 2 days now and I have been trying to find it on google but it's not listed anywhere.

Thanks
Take care,
Beckie

 

[stapel]

Can someone please give me the answer to (3-n)!

What do you mean by "the answer to"? The factorial isn't a question, that it could be "answered".

No one is answering me directly.

If you think this to be the case, then it might be good to review the replies you've received, because you have been given very direct answers.

It sounds like you guys are saying to do this: (3-n) - 1... then what do I multiply the next value or add it or subtract it?

When we say that you need to "multiply" the numbers together, yes, we mean "multiply", not add or subtract.

Eliz.

 

[Guest]

Please dont be rude.

I was just asking for the answer. I found on another website with this on it:

(n-1)! = (n-1)(n-2)(n-3)!

(the above equation works for all numbers of n)

Now that is an answer. This is what I wanted except I wanted it for:

(3-n)! = ?


Thanks
Take care,
Beckie

 

[pka]

Beckie, Please post the actual question!
I have taught this stuff for over thirty years.
But none of your questions make any sense to me.
If we could see the actual question, I think that we can straighten out the confusion implicit in all of your questions. If you really want help, why not post the actual wording of the problem?

 

[Guest]

I was trying to make it easier on you guys so you wouldn't have to try and solve this giant problem that would take a lot of time.

I just wanted to know how to do
(3-n)!

The type of answer I'm looking for is this:
(n-3)! = (n-3)(n-2)(n-1)!
except for the (3-n)!

Thanks
Beckie

 

[pka]

There is literally no way that the expression (3-n)! can occur in probability unless \(\displaystyle n \le 3\).
Therefore, I have concluded that the notation confuses you.
I have also concluded from your most resent reply that you really don’t want help!
So be it!

 

[Guest]

I do want help otherwise I wouldn't be here asking so many questions.

I've asked it so many different ways now. I don't know how to be any clearer?

Maybe this will help:

(I don't know how to do that math notation stuff, so here it is without it)

1- (C(48,n)/C(50,n)) >= 1/2

I need to know after this step how they cancel out the factorials?

Thanks
Beckie

 

[stapel]

Now that is an answer.

And that was the explanation we gave you for (n - 1)!.

This is what I wanted except I wanted it for: (3-n)! = ?

And this is a different (and highly unusual) expression, but we gave you the expansion for that, too.

I'm sorry if the answer isn't what you wanted, but that doesn't make the answer "rude".

I don't know how to be any clearer?

Providing the requested information would go a long way.

(C(48,n)/C(50,n))

Are you familiar with what the "m-choose-n" notation means?

I need to know after this step how they cancel out the factorials

If the work isn't shown in the book, try writing out the fractions yourself:

. . . . .\(\displaystyle \L \frac{\left(\frac{48!}{(48\, -\, n)!\, n!}\right)}{\left(\frac{50!}{(50\, -\, n)!\, n!}\right)}\)

The n!'s cancel out immediately:

. . . . .\(\displaystyle \L \frac{\left(\frac{48!}{(48\, -\, n)!}\right)}{\left(\frac{50!}{(50\, -\, n)!}\right)}\)

If you understand what factorials are and how they expand, then you will know how to simplify the (48!)/(50!) to get:

. . . . .\(\displaystyle \L \frac{\left(\frac{1}{(48\, -\, n)!}\right)}{\left(\frac{(49)(50)}{(50\, -\, n)!}\right)}\)

Now write out the symbolic expansion for (48 - n)! and (50 - n)!, and you should be able to get their final expression.

If you get stuck, please reply showing your work. Thank you.

Eliz.