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# Suppose n children are going to sit, in any order, on a row of n chair

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Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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23 Feb 2015, 03:24
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Suppose n children are going to sit, in any order, on a row of n chairs. What is the value of n?

(1) In this row, there are 5040 possible arrangements of the n children.
(2) If one child leaves, and one chair is removed from the row, there would be 4320 fewer arrangements of the remaining children.

Kudos for a correct solution.

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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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02 Mar 2015, 06:51
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Bunuel wrote:
Suppose n children are going to sit, in any order, on a row of n chairs. What is the value of n?

(1) In this row, there are 5040 possible arrangements of the n children.
(2) If one child leaves, and one chair is removed from the row, there would be 4320 fewer arrangements of the remaining children.

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

This is a question about permutations. Because this is Data Sufficiency, we don’t have to bother with the details of calculating n! —- it’s enough to know that if we know the value of n, we can figure out the value of n!, and vice versa!

Statement #1: this gives us n! = 5040 — from this we could figure out n. Statement #1, alone and by itself, is sufficient.

Statement #2: this tells us, essentially, n! – (n – 1)! = 4320. Well, if we went through list of factorials, there could be only one pair that is separated by exactly 4320. We don’t have to know what it is — it’s enough to know that this fact determines a unique value of n. Statement #2, alone and by itself, is sufficient.

(BTW, n = 7 if you’re curious!)
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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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23 Feb 2015, 03:40
1
statement 1
If n children sits in nchairs then arrangements is n!

Total arrangements is 5040. it equals to 7!.

It is sufficient.

Statement 2
By removing one children from n is( n-1) and one chair from n row is( n-1) than arrangements is n! - (n-1)!.
Arrangement is 4320 left. it means 7! - 6!.
5040 - 720= 4320.
It is also sufficient.

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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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04 Mar 2015, 23:56
Hi All,

There are going to be a few questions on Test Day that require the use of factorials. While it's not necessary to have these calculations memorized, if you DO have them memorized, then you'll likely be able to move much faster on these handful of questions.

2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040

You're not likely to have to deal with much higher than 7! as a 'solo' calculation. In a question such as this one, this knowledge can help you to quickly work through the prompt.

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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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14 Jun 2017, 13:30
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Statement 1 . Sufficient
there are 5040 possible arrangements of the n children And because n children can be arranged in n chairs in n!
thus n!=5040
Using prime factorisation I found n=7

Statement : 2 Sufficient
and if we remove one person and one chair then there would be 4320 scenario
n!-(n- 1)!=4320
Taking (n-1) common , (n-1).(n-1)!=4320
Using prime factorisation u can find (n-1) to be 6
6*6! = 4320

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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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14 May 2018, 07:17
Bunuel, chetan2u, EMPOWERgmatRichC, VeritasPrepKarishma

if $$n!-(n-1)!$$ = K (a constant integer)

Can the value of n be uniquely determined ?
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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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14 May 2018, 09:33
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gmatbusters wrote:
Bunuel, chetan2u, EMPOWERgmatRichC, VeritasPrepKarishma

if $$n!-(n-1)!$$ = K (a constant integer)

Can the value of n be uniquely determined ?

Note the pattern:
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720

The difference between consecutive factorials keeps increasing (obviously since every time you are multiplying it by the next greater number).
So there will be only one pair of consecutive values for any given constant.

Say K = 18. Then n = 4
Say K = 600. Then n = 6
and so on...
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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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14 May 2018, 10:02
1
Thank you for giving insights...

VeritasPrepKarishma wrote:

if $$n!-(n-1)!$$ = K (a constant integer)

Can the value of n be uniquely determined ?

Note the pattern:
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720

The difference between consecutive factorials keeps increasing (obviously since every time you are multiplying it by the next greater number).
So there will be only one pair of consecutive values for any given constant.

Say K = 18. Then n = 4
Say K = 600. Then n = 6
and so on...[/quote]
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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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14 May 2018, 14:21
1
Suppose n children are going to sit, in any order, on a row of n chairs. What is the value of n?

(1) In this row, there are 5040 possible arrangements of the n children.
(2) If one child leaves, and one chair is removed from the row, there would be 4320 fewer arrangements of the remaining children.

1) Would GMAT give 5041 as an answer? Nope, never seen it. Given it's a straight number, this will be sufficient.
2) Same logic applies. Would the difference of n! - (n-1)! be some random number? Nope; thus, this will be sufficient.

Without solving, I would wager on D.
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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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14 May 2018, 14:30
Explain u posted is helpful ! but do not understand why B can be right ? How can u calculate on exam with no calc?
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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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14 May 2018, 18:17
BeePen, please see the below explanation by veritasprepkarishma

if $$n!-(n-1)!$$ = K (a constant integer), n can be uniquely found

Note the pattern:
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720

The difference between consecutive factorials keeps increasing (obviously since every time you are multiplying it by the next greater number).

So there will be only one pair of consecutive values for any given constant.

Say K = 18. Then n = 4
Say K = 600. Then n = 6
and so on...[/quote]
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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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18 Jul 2018, 00:01
Bunuel niks18 chetan2u pushpitkc

Quote:
Suppose n children are going to sit, in any order, on a row of n chairs. What is the value of n?

(2) If one child leaves, and one chair is removed from the row, there would be 4320 fewer arrangements of the remaining children.

Can you please explain St 2: How we interpret : n! – (n – 1)! = 4320
when we have same no of children and chair and start and now we removed each of one chair and one children.
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Re: Suppose n children are going to sit, in any order, on a row of n chair  [#permalink]

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18 Jul 2018, 03:14
1
Bunuel niks18 chetan2u pushpitkc

Quote:
Suppose n children are going to sit, in any order, on a row of n chairs. What is the value of n?

(2) If one child leaves, and one chair is removed from the row, there would be 4320 fewer arrangements of the remaining children.

Can you please explain St 2: How we interpret: n! – (n – 1)! = 4320
when we have same no of children and chair and start and now we removed each of one chair and one children.

The 'n' children can be arranged in 'n' chairs is n! ways
After 1 chair is removed, the 'n-1' children can be arranged in (n-1)! ways

Statement 2 says that if one child leaves and one chair is removed

To understand this clearly, let's consider n = 3.

So, how can we seat 3 children in 3 chairs -
3 ways of seating one of the 3 children in the first chair
2 ways of seating one of the 2 children in the second chair
1 ways of seating one child in the third chair (3*2*1), which is 3!

Similarly, we can seat 2 children in 2 chairs as (2*1), which is 2!

Therefore, the total number of ways must be n! - (n-1)!

Hope this helps you!
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Re: Suppose n children are going to sit, in any order, on a row of n chair &nbs [#permalink] 18 Jul 2018, 03:14
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