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Difficulty: 505-555 Level,    Algebra,                   
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
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Bunuel wrote:
The expression n! is defined as the product of the integers from 1 through n. If p is the product of the integers from 100 through 299 and q is the product of the integers from 200 through 299, which of the following is equal to p/q?

A. 99!
B. 199!
C. 199!/99!
D. 299!/99!
E. 299!/199!


The quotient p/q is the product of the integers from 100 to 199, inclusive, since the product of the integers from 200 to 299, inclusive, appears in both p and q and thus it gets canceled out.

We can consider the product of the integers from 100 to 199, inclusive, as the product of the integers from 1 to 199, inclusive, divided by the product of the integers from 1 to 99, inclusive. That is,

p/q = the product of the integers from 100 to 199, inclusive = 199!/99!

Answer: C
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
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Bunuel wrote:
The expression n! is defined as the product of the integers from 1 through n. If p is the product of the integers from 100 through 299 and q is the product of the integers from 200 through 299, which of the following is equal to p/q?

A. 99!
B. 199!
C. 199!/99!
D. 299!/99!
E. 299!/199!

NEW question from GMAT® Quantitative Review 2019


(PS01656)


Given,
p=100*101*102*.......................*298*299 and
q=200*201*202*.......................*298*299
Thus, \(\frac{p}{q}\)=\(\frac{101*102*103*.......................*298*299}{200*201*202*.......................*298*299}\)=100*101*102*--------------*198*199=\(\frac{1*2*3*.................*198*199}{1*2*3*...............*98*99}\)=\(\frac{199!}{99!}\)

Ans. (C)
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
Bunuel wrote:
The expression n! is defined as the product of the integers from 1 through n. If p is the product of the integers from 100 through 299 and q is the product of the integers from 200 through 299, which of the following is equal to p/q?

A. 99!
B. 199!
C. 199!/99!
D. 299!/99!
E. 299!/199!

NEW question from GMAT® Quantitative Review 2019


(PS01656)


Given:
1. The expression n! is defined as the product of the integers from 1 through n.
2, p is the product of the integers from 100 through 299
3. q is the product of the integers from 200 through 299

Asked: Which of the following is equal to p/q?

\(p = \frac{299!}{99!}\)

\(q = \frac{299!}{199!}\)

\(\frac{p}{q} = \frac{ \frac{299!}{99!} }{ \frac{299!}{199!}}\)


\(\frac{p}{q} = \frac{199!}{99!}\)



IMO C
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
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can anyone explain this, why isn't the answer 199!
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
Did not understand the question, request if someone could kindly elaborate

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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
Hi Stutidutta1

Quote:
Did not understand the question, request if someone could kindly elaborate

Could you elaborate what did you not understand? See this post to help yourself.

You can give it a shot to attempt anything (even a little) you understood and and members/ experts can take it up from there.
Happy study prep :)
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
Kinshook wrote:
Bunuel wrote:
The expression n! is defined as the product of the integers from 1 through n. If p is the product of the integers from 100 through 299 and q is the product of the integers from 200 through 299, which of the following is equal to p/q?

A. 99!
B. 199!
C. 199!/99!
D. 299!/99!
E. 299!/199!

NEW question from GMAT® Quantitative Review 2019


(PS01656)


Given:
1. The expression n! is defined as the product of the integers from 1 through n.
2, p is the product of the integers from 100 through 299
3. q is the product of the integers from 200 through 299

Asked: Which of the following is equal to p/q?

\(p = \frac{299!}{99!}\)

\(q = \frac{299!}{199!}\)

\(\frac{p}{q} = \frac{ \frac{299!}{99!} }{ \frac{299!}{199!}}\)


\(\frac{p}{q} = \frac{199!}{99!}\)



IMO C



Can you explain how & why do we get 99! in the denominator for \(p = \frac{299!}{99!}\) ?
Thanks!
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
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trivedipratik wrote:
Kinshook wrote:
Bunuel wrote:
The expression n! is defined as the product of the integers from 1 through n. If p is the product of the integers from 100 through 299 and q is the product of the integers from 200 through 299, which of the following is equal to p/q?

A. 99!
B. 199!
C. 199!/99!
D. 299!/99!
E. 299!/199!

NEW question from GMAT® Quantitative Review 2019


(PS01656)


Given:
1. The expression n! is defined as the product of the integers from 1 through n.
2, p is the product of the integers from 100 through 299
3. q is the product of the integers from 200 through 299

Asked: Which of the following is equal to p/q?

\(p = \frac{299!}{99!}\)

\(q = \frac{299!}{199!}\)

\(\frac{p}{q} = \frac{ \frac{299!}{99!} }{ \frac{299!}{199!}}\)


\(\frac{p}{q} = \frac{199!}{99!}\)



IMO C



Can you explain how & why do we get 99! in the denominator for \(p = \frac{299!}{99!}\) ?
Thanks!


trivedipratik Rhl5084

so initially we have

\(\frac{p}{q} =\frac{ (100*101 ....200*299)}{ (200*201...*299)} \)

in the denominator and numerator 200*201...*299 is canceled out.

so you are left with factorial 100*101 ....*199

notice it starts from 100 and not from 1. If it started from 1 then answer choice would be B i.e 199!

but since it starts from 100 you need to cancel out 99! ie \(100*101 ....*199 =\frac{ (1*2 ....*199)}{(99!)}\)

:)
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
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The expression n! is defined as the product of the integers from 1 through n. If p is the product of the integers from 100 through 299 and q is the product of the integers from 200 through 299, which of the following is equal to p/q?

My approach:

\(p = \frac{299!}{99!}\)

\(q = \frac{299!}{199!}\)

\(\frac{p}{q}= 100*101*102...*199\)

This can be written as \(\frac{199!}{99!}\)

Answer is C.
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
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Bunuel wrote:
The expression n! is defined as the product of the integers from 1 through n. If p is the product of the integers from 100 through 299 and q is the product of the integers from 200 through 299, which of the following is equal to p/q?

A. 99!
B. 199!
C. 199!/99!
D. 299!/99!
E. 299!/199!

NEW question from GMAT® Quantitative Review 2019


(PS01656)



\(\frac{100*101*.....*199*200*.....*299}{200*201*.....*299}\)

=100*101*.....*199

Answer choices are given FACTORIAL FORM.
A. 99! is bellow 100
B. 199! is up to 199, but it includes more numbers from 1*2*.....99
C. This option Reach up to 199! and cancel out excess values by dividing 199! by 99!

So, the answer is C
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
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Hi,

I used a totally different approach from the above ones, I checked it with smaller numbers and it works

p= from 100 to 299 and q = from 200 to 299
p : 299!-100! = 199!
q: 299!-200! = 99!
p/q=199!/99!

Answer C)
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
manvi17 wrote:
can anyone explain this, why isn't the answer 199!



simply because we're left with 100*101*102...*199. notice, this does not start with a 1, but with 100, and as you know, 199! is 1*2*3*4......*198*199. so, that why we divide it by 99!
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
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P = 299!/99!
Q = 299!/199!
Therefore P/Q = 199!/99!

Hope It helps :)
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
n! = 1*...*n
p!=100*..*299
q!=200*...*299
p/q =(299-100)!/(299-200)! = 199!/99! (C)
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
I was trying to solve this in the following manner which is wrong.
Please help why this approach is wrong.

P/Q= 299!-99!/299!-199!

Where am I wrong?

Posted from my mobile device
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
Expert Reply
 
Iashish.sharma wrote:
The expression n! is defined as the product of the integers from 1 through n. If p is the product of the integers from 100 through 299 and q is the product of the integers from 200 through 299, which of the following is equal to p/q?

A. 99!
B. 199!
C. 199!/99!
D. 299!/99!
E. 299!/199!

I was trying to solve this in the following manner which is wrong.
Please help why this approach is wrong.

P/Q= 299!-99!/299!-199!

Where am I wrong?

Posted from my mobile device

­
We are given that \(p = 100*101*...*299\), which can be written as \(\frac{299!}{99!}\), and that does not equalt to \(299! - 99!\).

Similarly, \(q = 200*201*...*299\), which can be written as \(\frac{299!}{199!}\), and that does not equalt to \(299! - 199!\).­
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Re: The expression n! is defined as the product of the integers from 1 thr [#permalink]
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