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In how many different ways can 8 people be seated in a room...
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11 Aug 2016, 01:42
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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 Please explain in detail your answer so that we can all follow it.
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In how many different ways can 8 people be seated in a room...
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11 Aug 2016, 01:47
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
Please explain in detail your answer so that we can all follow it. \(10P8 = \frac{10!}{(108)!} = 1,814,400\), choosing 8 out of 10, when order matters. Answer: D.
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Re: In how many different ways can 8 people be seated in a room...
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11 Aug 2016, 03:13
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
Please explain in detail your answer so that we can all follow it. 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...
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11 Aug 2016, 03:21
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
Please explain in detail your answer so that we can all follow it. 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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In how many different ways can 8 people be seated in a room...
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11 Aug 2016, 03:28
chisicheiWhen 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...
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11 Aug 2016, 10:23
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...
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11 Aug 2016, 12:00
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
Please explain in detail your answer so that we can all follow it. 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...
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03 Feb 2017, 08:45
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...
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16 Jun 2017, 23:51
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
Please explain in detail your answer so that we can all follow it. Here's two more ways to think about this one. 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 \(\frac{10!}{2}\) Answer (D) 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. \(10C2 * 8!\) Answer (D)
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Re: In how many different ways can 8 people be seated in a room...
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17 Jun 2017, 03:15
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...
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03 Jul 2017, 02:07
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 rewritten 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...
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11 Sep 2018, 23:00
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
Please explain in detail your answer so that we can all follow it. 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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