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# In how many different ways can 8 people be seated in a room...

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Manager
Joined: 24 May 2016
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11 Aug 2016, 02:42
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Difficulty:

55% (hard)

Question Stats:

64% (02:02) correct 36% (01:51) wrong based on 193 sessions

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In how many different ways can 8 people be seated in a room with 10 chairs?

A) 40,320
B) 181,440
C) 403,200
D) 1,814,400
E) 3,628,800

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Posts: 53770
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11 Aug 2016, 02:47
1
EBITDA wrote:
In how many different ways can 8 people be seated in a room with 10 chairs?

A) 40,320
B) 181,440
C) 403,200
D) 1,814,400
E) 3,628,800

$$10P8 = \frac{10!}{(10-8)!} = 1,814,400$$, choosing 8 out of 10, when order matters.

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Re: In how many different ways can 8 people be seated in a room...  [#permalink]

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11 Aug 2016, 04:13
EBITDA wrote:
In how many different ways can 8 people be seated in a room with 10 chairs?

A) 40,320
B) 181,440
C) 403,200
D) 1,814,400
E) 3,628,800

Order doesn't matter
With 10 chairs and only 8 to seat
8!- Number of ways the 8 can seat on the chair
10C2 - Number of ways the 2 empty sits can be vary with the 8 seated

8! * 10C2 = 1,814,400 Answer - D
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Re: In how many different ways can 8 people be seated in a room...  [#permalink]

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11 Aug 2016, 04:21
1
EBITDA wrote:
In how many different ways can 8 people be seated in a room with 10 chairs?

A) 40,320
B) 181,440
C) 403,200
D) 1,814,400
E) 3,628,800

Hi,
The logic is

Let the first person sit in one out of 10...
So next will have 9 to choose from and next 8 and so on..
So total ways 10*9*8*7*6*5*4*3 which is same as 10P8...
D
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11 Aug 2016, 04:28
chisichei

When you say that order does not matter, I presume that you mean that the order among the empty seats does not matter. Am I right?
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Re: In how many different ways can 8 people be seated in a room...  [#permalink]

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11 Aug 2016, 11:23
Order doesnot matter

no of ways of selecting 8 chairs of 10 chairs =10C8 ways
In these 8 chairs 8 people can be seated in 8! ways

SO total =10C8 *8!
which is also equal to 10P8
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Re: In how many different ways can 8 people be seated in a room...  [#permalink]

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11 Aug 2016, 13:00
Top Contributor
chisichei wrote:
EBITDA wrote:
In how many different ways can 8 people be seated in a room with 10 chairs?

A) 40,320
B) 181,440
C) 403,200
D) 1,814,400
E) 3,628,800

Order doesn't matter
With 10 chairs and only 8 to seat
8!- Number of ways the 8 can seat on the chair
10C2 - Number of ways the 2 empty sits can be vary with the 8 seated

8! * 10C2 = 1,814,400 Answer - D

Nice Solution

But I believe order matters and order among empty seat also matters here and you made the solution considering all orders for this problem.

Did I missed something?
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Re: In how many different ways can 8 people be seated in a room...  [#permalink]

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03 Feb 2017, 09:45
In how many different ways can 12 people be seated in a room with 10 chairs?
Is it 12C10*10!?
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In how many different ways can 8 people be seated in a room...  [#permalink]

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17 Jun 2017, 00:51
EBITDA wrote:
In how many different ways can 8 people be seated in a room with 10 chairs?

A) 40,320
B) 181,440
C) 403,200
D) 1,814,400
E) 3,628,800

1.
There are in total 10 things to be arranged, i.e. 8 people and 2 gaps(which are similar).

Number of ways of arranging 10 things out of which 2 are similar

$$\frac{10!}{2}$$

2.
as there are 10 chairs and 8 people, we will have to first choose which 8 chairs to fill. After we have chosen our chairs, we can arrange the people.

$$10C8 * 8!$$

This can also be imagined in another way, we have to choose 2 chairs to leave empty. And then we have to arrange the people in the remaining 8 chairs.

$$10C2 * 8!$$

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Re: In how many different ways can 8 people be seated in a room...  [#permalink]

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17 Jun 2017, 04:15
1
It is very clearly 10C8*8!= 10*9*8*7*6*5*4*3

We don't need to calculate the above. Here's the trick

1. How many number of zeros can this have?
Number of zeros= highest power of 10, which is equal to 2
A and B eliminated

The remaining three choices have different digits before zero, thus if we can figure out the digit we will get the answer.
The digit before zero is 4. Thus D is the answer

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Re: In how many different ways can 8 people be seated in a room...  [#permalink]

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03 Jul 2017, 03:07
I'm not sure if the calculation portion is within the scope of the GMAT but given the spread of the numbers, it is definitely doable.

The trick is to get everything to the power of base 10. I won't go through the entire exercise but I will start with this example:
10!/2! = 10*9*8*7*6*5*4*3*2 can be re-written as:
10*10*9*8*7*6*_*4*3*_*
Notice the 2 and 5 missing - they were actually multiplied to get another 10. Here is one more (less straightforward step):
10*10*10*10*1.12*9*6*2*3
In the step I just wrote out, we multiplied 7 * 8 * 2(from the 4) to get 112 which we rewrote as 1.12 * 100(10*10)
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Re: In how many different ways can 8 people be seated in a room...  [#permalink]

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12 Sep 2018, 00:00
EBITDA wrote:
In how many different ways can 8 people be seated in a room with 10 chairs?

A) 40,320
B) 181,440
C) 403,200
D) 1,814,400
E) 3,628,800

Selecting 8 seats on which 8 people would be seated - 10C8.

This can be arranged in 8! ways.

Answer = $$10C8 * 8! = 1,814,400$$
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Re: In how many different ways can 8 people be seated in a room...   [#permalink] 12 Sep 2018, 00:00
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