The expression n! is defined as the product of the integers from 1 thr

Question

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)

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

(PS01656)

Princ · Accepted Answer

OA:C

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

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

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

\frac{p}{q}=\frac{{299!}}{{99!}}*\frac{{199!}}{{299!}}=\frac{{199!}}{{99!}}

dave13 · Answer

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!)}

dvishal387 · Answer

200 through 299 will cancel out leaving 100 through 199.
It can be written as 199!/99!.
C should be the ans.

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ScottTargetTestPrep · Answer

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

PKN · Answer

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)

Kinshook · Answer

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

manvi17 · Answer

can anyone explain this, why isn't the answer 199!

Stutidutta1 · Answer

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

Posted from my mobile device

adkikani · Answer

Hi Stutidutta1 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

trivedipratik · Answer

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

Basshead · Answer

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)

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.

MHIKER · Answer

\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

Sabby · Answer

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)

ueh55406 · Answer

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!

1610ayush · Answer

P = 299!/99! Q = 299!/199! Therefore P/Q = 199!/99! Hope It helps

Kritika09 · Answer

n! = 1*...*n
p!=100*..*299
q!=200*...*299
p/q =(299-100)!/(299-200)! = 199!/99! (C)

Iashish.sharma · Answer

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

Bunuel · Answer

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! .

MUB · Answer

which math topic covers this kind of questions in gmat focus edition

The expression n! is defined as the product of the integers from 1 thr

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