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# If n > 2, then what is the value of (n + 1)!/(n - 2)! ?

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Joined: 02 Sep 2009
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If n > 2, then what is the value of (n + 1)!/(n - 2)! ?  [#permalink]

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Updated on: 04 Mar 2019, 01:07
00:00

Difficulty:

15% (low)

Question Stats:

81% (01:19) correct 19% (01:16) wrong based on 31 sessions

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If n > 2, then what is the value of $$\frac{(n + 1)!}{(n - 2)!}$$ ?

A. 0
B. n!
C. $$n^3 – n$$
D. $$n^3 − 4n$$
E. (2n)!

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Originally posted by Bunuel on 04 Mar 2019, 00:00.
Last edited by carcass on 04 Mar 2019, 01:07, edited 1 time in total.
Edited by Carcass
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Re: If n > 2, then what is the value of (n + 1)!/(n - 2)! ?  [#permalink]

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04 Mar 2019, 00:44

Solution

Given:
• The value of n is greater than 2

To find:
• The value of the expression $$\frac{(n+1)!}{(n-2)!}$$

Approach and Working:
We can write the number (n + 1)! in the following manner:
• (n + 1)! = n! * (n + 1) = (n – 1)! * n * (n + 1) = (n – 2)! * (n – 1) * n * (n + 1)

• Therefore, the value of the given expression = $$(n – 2)! * (n – 1) * n * (n + 1) * \frac{1}{(n – 2)!} = (n – 1) * n * (n + 1) = n (n^2 – 1) = n^3 – n$$

Hence, the correct answer is option C.

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Re: If n > 2, then what is the value of (n + 1)!/(n - 2)! ?  [#permalink]

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04 Mar 2019, 01:03
Bunuel wrote:
If n > 2, then what is the value of $$\frac{(n + 1)!}{(n - 2)!}$$ ?

A. 0
B. n!
C. n^3 – n
D. n^3 − 4n
E. (2n)!

take n = 3
so we get
$$\frac{(n + 1)!}{(n - 2)!}$$
= 24
which is equal to option C ; n^3 – n
IMO C
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Re: If n > 2, then what is the value of (n + 1)!/(n - 2)! ?  [#permalink]

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04 Mar 2019, 02:37
Bunuel wrote:
If n > 2, then what is the value of $$\frac{(n + 1)!}{(n - 2)!}$$ ?

A. 0
B. n!
C. $$n^3 – n$$
D. $$n^3 − 4n$$
E. (2n)!

let n be 3.

$$\frac{(n + 1)!}{(n - 2)!}$$

=$$\frac{(3 + 1)!}{(3 - 2)!}$$

=$$\frac{4!}{1!}$$

= 24.

Option C:

$$n^3 - n$$

= $$3^3 - 3$$

= 24.

Thus , C is the correct answer.

For any value of n>2, This system will work the same as above.
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Re: If n > 2, then what is the value of (n + 1)!/(n - 2)! ?  [#permalink]

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06 Mar 2019, 19:58
Bunuel wrote:
If n > 2, then what is the value of $$\frac{(n + 1)!}{(n - 2)!}$$ ?

A. 0
B. n!
C. $$n^3 – n$$
D. $$n^3 − 4n$$
E. (2n)!

We see that (n + 1)! = (n + 1)(n)(n - 1)(n - 2)(n - 3)(n - 4)...

We see that (n - 2)! = = (n - 2)(n - 3)(n - 4)...

So we see that we are left with (n + 1)(n)(n - 1) = n(n^2 - 1) = n^3 - n.

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Re: If n > 2, then what is the value of (n + 1)!/(n - 2)! ?   [#permalink] 06 Mar 2019, 19:58
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