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Re: Which of the following is an integer?<#permalink>18 Aug 2009, 19:30

Did a search, can not find anypoint on this one (the question stem is pretty vague). Would appreciate some enlightenment on the shortreduced I"m missing!

**Which of the adhering to is an integer?I. 12! / 6!II. 12! / 8!III. 12! / 7!5!A) I onlyB) II onlyC) III onlyD) I and II onlyE) I, II, and also IIISource: GMAT Prep Exam. Equipment : E**

To me the question looks right forward.we understand 12! is multiple of all numbers from 1 to 12, so separating it by 6! or 8! is pretty a lot sure an integer.now coming to the 3rd options 12! by 5! leaves 12*11*10*9*8*7*6/1*2*3*4*5*6*7 which results in integer, so the answe E is right

Tbelow is no shortreduced persay...

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To me the question looks right forward.we understand 12! is multiple of all numbers from 1 to 12, so separating it by 6! or 8! is pretty a lot sure an integer.now coming to the 3rd options 12! by 5! leaves 12*11*10*9*8*7*6/1*2*3*4*5*6*7 which results in integer, so the answe E is right

Tbelow is no shortreduced persay...

You are watching: Which of the following is an integer

**Clear E.12!/6! = 12*11*10*9*8*7*6!/6! = Integer12!/8! = 12*11*10*9*8!/8! = Integer12!/7!*5! = 12*11*10*9*8/5*4*3*2*1 = integerSo all 3 are correct.**

Every factorial is divisible by every smaller sized factorial, so 12!/6! and also 12!/8! are integers. 12!/(7!*5!) is the variety of means to choose 7 things from a team of 12 if order is irrelevant, so it should additionally be an integer.Every factorial is divisible by every smaller sized factorial, so 12!/6! and also 12!/8! are integers. 12!/(7!*5!) is the variety of means to choose 7 things from a team of 12 if order is irrelevant, so it should additionally be an integer.

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case 1 and also situation 2 are straightforward however situation 3 is harder. for instance 3, we need to write down all the determinants.

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case 1 and also situation 2 are straightforward however situation 3 is harder. for instance 3, we need to write down all the determinants.

Hi All,

**Beyond the arithmetic that sdrandom1 stated, these calculations autumn into some patterns that you are likely to see (in some form) on Test Day. 12!/6! When handling INDIVIDUAL factorials, if the factorial in the numerator is GREATER than the factorial in the denominator, then you will end up with an integer. Here, since 12 > 6, 12!/6! will certainly simplify to an integer. 12!/7!5! You could not have studied the Combination Formula yet, however this calculation is what you would certainly end up through as soon as answering the question "just how many kind of various combicountries of 7 people deserve to you develop from a team of 12 people?" In these types of "Combination" calculations, the number of groups will certainly always be an integer, so 12!/7!5! will certainly simplify to an integer. Final Answer:**

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Which of the adhering to is an integer?empowergmat.com The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★

Which of the adhering to is an integer?

**I. 12! / 6!II. 12! / 8!III. 12! / 7!5!A) I onlyB) II onlyC) III onlyD) I and II onlyE) I, II, and III**

For III, I went through and crossed out whatever that"s diminished to visualize...12*11*102*93*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 * 5*4*3*2*1So you"re left with 12*11*2*3 (from taking 5 and also 3 from 10 and 9) = 792. Makes feeling.ScottTargetTestPrep As a rule, you said m = n + p, then m!/(n!p!). When does the factorial in the denominator make the outcome not an integer? Am I ideal to think this hregarding perform through the number of prime factors?Because 12!/8!5! and 12!/9!5! are additionally integers, while 12!/10!5! isn"t. If we had 11*2^3*3^2 left with 7! on the bottom then 8! would be 2^3 and also 9! would be 3^2 both of which we can still mitigate... while 10! provides us a aspect of 5 which we do not have.

Which of the adhering to is an integer?For III, I went through and crossed out whatever that"s diminished to visualize...12*11*102*93*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 * 5*4*3*2*1So you"re left with 12*11*2*3 (from taking 5 and also 3 from 10 and 9) = 792. Makes feeling.ScottTargetTestPrep As a rule, you said m = n + p, then m!/(n!p!). When does the factorial in the denominator make the outcome not an integer? Am I ideal to think this hregarding perform through the number of prime factors?Because 12!/8!5! and 12!/9!5! are additionally integers, while 12!/10!5! isn"t. If we had 11*2^3*3^2 left with 7! on the bottom then 8! would be 2^3 and also 9! would be 3^2 both of which we can still mitigate... while 10! provides us a aspect of 5 which we do not have.

Which of the adhering to is an integer?

**I. 12! / 6!II. 12! / 8!III. 12! / 7!5!A) I onlyB) II onlyC) III onlyD) I and II onlyE) I, II, and III**

For III, I went through and crossed out every little thing that"s decreased to visualize...12*11*102*93*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 * 5*4*3*2*1So you"re left through 12*11*2*3 (from taking 5 and also 3 from 10 and 9) = 792. Makes sense.ScottTargetTestPrep As a preeminence, you shelp m = n + p, then m!/(n!p!). When does the factorial in the denominator make the outcome not an integer? Am I best to think this has to execute via the number of prime factors?Because 12!/8!5! and also 12!/9!5! are also integers, while 12!/10!5! isn"t. If we had 11*2^3*3^2 left via 7! on the bottom then 8! would be 2^3 and 9! would certainly be 3^2 both of which we can still mitigate... while 10! gives us a aspect of 5 which we don"t have.

TargetTestPrep.comFor III, I went through and crossed out every little thing that"s decreased to visualize...12*11*102*93*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 * 5*4*3*2*1So you"re left through 12*11*2*3 (from taking 5 and also 3 from 10 and 9) = 792. Makes sense.ScottTargetTestPrep As a preeminence, you shelp m = n + p, then m!/(n!p!). When does the factorial in the denominator make the outcome not an integer? Am I best to think this has to execute via the number of prime factors?Because 12!/8!5! and also 12!/9!5! are also integers, while 12!/10!5! isn"t. If we had 11*2^3*3^2 left via 7! on the bottom then 8! would be 2^3 and 9! would certainly be 3^2 both of which we can still mitigate... while 10! gives us a aspect of 5 which we don"t have.

TargetTestPrep.com

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This question gives you an excellent chance to testimonial factorials, and once you answer this question you"ll be well-prepared for various other inquiries that involve factorials.

**Here"s an explanation on YouTubehttps://www.youtube.com/watch?v=U7kymGI ... PfYZOmjn_0Answer (E)**

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