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((n-1)!+n!+(n+1)!)/n! Updated on: 18 Feb 2016, 15:26
Originally posted by jahanafsana on 08 Mar 2015, 18:25.
Last edited by ENGRTOMBA2018 on 18 Feb 2016, 15:26, edited 1 time in total.
Reformatted the question and added the OA.
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\(\frac{(n-1)!+n!+(n+1)!}{n!}\)

Which of the following values of n will result between 16 and 17 for the equation?

A. 14
B. 15
C. 16
D. 17
E. 18

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If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.
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A
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((n-1)!+n!+(n+1)!)/n! 09 Mar 2015, 23:50
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jahanafsana
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18


\(\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)\)

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)
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((n-1)!+n!+(n+1)!)/n! 08 Mar 2015, 18:41
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jahanafsana
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18

hi jahanafsana,
it depends how you have reduced the equation...
1) first is you open each term .... find common terms... get an equation and then substitute values..
2) second , which may be slightly less time consuming.....
\(\frac{((n-1)!+n!+(n+1)!)}{n!}\)
=\(\frac{(n-1)!}{n!}\)+\(\frac{n!}{n!}\)+\(\frac{(n+1)!}{n!}\)
=\(\frac{1}{n}\)+1+n+1.... now you can easily see n+2 should be greater than or equal to 16.... 14 is the value...
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CrackverbalGMAT
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((n-1)!+n!+(n+1)!)/n! 11 Mar 2015, 23:58
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Easiest way is simplify the equation and then check the answer choice.
(n-1)! +n! + (n+1)! /n! = (I/n) +1+ (n+1)
(n+2) + (I/n) is 16 or more than that so only option is 14.
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((n-1)!+n!+(n+1)!)/n! 12 Mar 2015, 09:44
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CrackVerbalGMAT
Easiest way is simplify the equation and then check the answer choice.
(n-1)! +n! + (n+1)! /n! = (I/n) +1+ (n+1)
(n+2) + (I/n) is 16 or more than that so only option is 14.


Thank you for your reply. It helped.
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((n-1)!+n!+(n+1)!)/n! 12 Mar 2015, 09:46
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CrackVerbalGMAT
Easiest way is simplify the equation and then check the answer choice.
(n-1)! +n! + (n+1)! /n! = (I/n) +1+ (n+1)
(n+2) + (I/n) is 16 or more than that so only option is 14.

Thanks a lot for your detail reply. It is clear now.
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((n-1)!+n!+(n+1)!)/n! 18 Feb 2016, 13:51
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VeritasPrepKarishma
jahanafsana
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18


\(\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)\)

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)
Could you show how you simplified this part in steps? I seem to get a different answer.
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((n-1)!+n!+(n+1)!)/n! 18 Feb 2016, 15:20
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VeritasPrepKarishma
jahanafsana
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18


\(\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)\)

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)
Could you show how you simplified this part in steps? I seem to get a different answer.

You can adopt 2 methods to verify: either substitute n=3 and cross check. For algebraic method look below:

Realize that (n+1)! = (n+1)*n*(n-1)(n-2)... = (n+1)*n!

Similarly, n!=n*(n-1)! ---> (n-1)!= n!/n

Thus, \(\frac{(n - 1)! + n! + (n + 1)!}{n!} =\frac{(n!/n)+n!+(n+1)n!}{n!}\) = (cancelling n! from both the numerator and denominator you get) \(\frac{1}{n}+1+(n+1)\)

Hope this helps.
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((n-1)!+n!+(n+1)!)/n! 18 Feb 2016, 20:46
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jahanafsana
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18


\(\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)\)

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)
Could you show how you simplified this part in steps? I seem to get a different answer.

\(\frac{((n-1)!+n!+(n+1)!)}{n!}\)
=\(\frac{(n-1)!}{n!}\)+\(\frac{n!}{n!}\)+\(\frac{(n+1)!}{n!}\)
=\(\frac{1}{n}\)+1+n+1....
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((n-1)!+n!+(n+1)!)/n! 18 Feb 2016, 20:59
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VeritasPrepKarishma
jahanafsana
(n-1)!+n!+(n+1)!
n!

Which of the following values of n will result between 16 and 17 for the equation?

If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

a. 14
b. 15
c. 16
d. 17
e. 18


\(\frac{(n - 1)! + n! + (n + 1)!}{n!} = \frac{1}{n} + 1 + (n+1)\)

Note that out of the options, whatever value you give n, 1/n will be a very small term, much smaller than 1.
In rest of the expression, you have 1 + (n + 1) which you want to be 16. So n must be 14.

Answer (A)
Could you show how you simplified this part in steps? I seem to get a different answer.

Note that
\((n-1)! = 1*2*3*...*(n-1)\)
\(n! = 1*2*3*...*(n-1)*n\)

So n! just has an extra n in it.

When you divide (n-1)! by n!, you are just left with an n in the denominator.

Similarly,
\(n! = 1*2*3*...*(n-1)*n\)
\((n+1)! = 1*2*3*...*(n-1)*n*(n+1)\)
So (n+1)! only has an extra (n+1).

When you divide (n+1)! by n!, you are left with (n+1) in the denominator.

\(\frac{(n - 1)! + n! + (n + 1)!}{n!}\)

\(= \frac{(n - 1)!}{n!} + \frac{n!}{n!} + \frac{(n+1)!}{n!}\)

\(= \frac{1}{n} + 1 + (n+1)\)
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((n-1)!+n!+(n+1)!)/n! 19 Feb 2016, 13:01
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VeritasPrepKarishma Thank you. Your explanation has been quite useful.
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((n-1)!+n!+(n+1)!)/n! 11 May 2017, 02:49
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jahanafsana
\(\frac{(n-1)!+n!+(n+1)!}{n!}\)

Which of the following values of n will result between 16 and 17 for the equation?

A. 14
B. 15
C. 16
D. 17
E. 18

Show Spoiler
If I factor (n-1) and reduce the equation and then substitute the answer choices then result becomes a. But is there a shortcut to solving this.

\(\frac{(n-1)!+n!+(n+1)!}{n!}\)
=1/n+1+n+1
=n+2+(1/n)
16<n+2+1/n<17
->14<n+1/n<15
n=14
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Answer A.
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((n-1)!+n!+(n+1)!)/n! 15 Nov 2021, 07:59
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jahanafsana
\(\frac{(n-1)!+n!+(n+1)!}{n!}\)

Which of the following values of n will result between 16 and 17 for the equation?

A. 14
B. 15
C. 16
D. 17
E. 18


Useful fraction property: \(\frac{a+b+c}{d}=\frac{a}{d}+\frac{b}{d}+\frac{c}{d}\)

Apply the property to get: \(\frac{(n-1)!+n!+(n+1)!}{n!}=\frac{(n-1)!}{n!}+\frac{n!}{n!}+\frac{(n+1)!}{n!}\)

\(=\frac{1}{n}+1+(n+1)\)

\(=n + 2 + \frac{1}{n}\)

From the answer choices, we can see that the value of n must be an integer from 14 to 18 inclusive, which means the fraction \(\frac{1}{n}\) must be between 0 and 1.

This means the expression \(n + 2 + \frac{1}{n}\) will be between 16 and 17 when \(n = 14\)

Answer: A
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