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# A double decked bus can accommodate 110 passengers, 50 in the upper de

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Math Expert
Joined: 02 Sep 2009
Posts: 59573
A double decked bus can accommodate 110 passengers, 50 in the upper de  [#permalink]

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12 Nov 2019, 02:40
00:00

Difficulty:

55% (hard)

Question Stats:

30% (01:34) correct 70% (02:16) wrong based on 30 sessions

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A double decked bus can accommodate 110 passengers, 50 in the upper deck and 60 in the lower deck.In how many ways can the passengers be accommodated if 15 refuse to be in the upper deck while 10 others refuse to be in the lower deck?

A. $$\frac{85!50!60!}{40!45!}$$

B. $$\frac{85!}{40!45!}$$

C. $$\frac{110!}{50!60!}$$

D. $$\frac{110!50!60!}{40!45!}$$

E. $$\frac{110!}{40!45!}$$

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Re: A double decked bus can accommodate 110 passengers, 50 in the upper de  [#permalink]

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12 Nov 2019, 02:54
1
total seats = 110
UP= 50
LD= 60
10 refuse LD and 15 refuse UP
60P15 *50P10* 85!
IMO A
$$\frac{85!50!60!}{40!45!}$$

Bunuel wrote:
A double decked bus can accommodate 110 passengers, 50 in the upper deck and 60 in the lower deck.In how many ways can the passengers be accommodated if 15 refuse to be in the upper deck while 10 others refuse to be in the lower deck?

A. $$\frac{85!50!60!}{40!45!}$$

B. $$\frac{85!}{40!45!}$$

C. $$\frac{110!}{50!60!}$$

D. $$\frac{110!50!60!}{40!45!}$$

E. $$\frac{110!}{40!45!}$$

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3158
Re: A double decked bus can accommodate 110 passengers, 50 in the upper de  [#permalink]

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13 Nov 2019, 03:46

Solution

Given
• A double-decked bus can accommodate 110 passengers.
o 50 passengers in the upper deck and 60 passengers in the lower deck.
• 15 passengers refuse to be in the upper deck and 10 other passengers refuse to be in the lower deck.

To find
• The number of ways can the passengers be accommodated in the bus
.

Approach and Working out

• Total number of arrangements= Selecting and arranging 15 passengers who refuse to be in the upper deck × Selecting and arranging 10 passengers who refuse to be in the upper deck × Arranging remaining 85 passengers
• = 60C15 × 15! × 50C10 × 10! × 85!
• = $$\frac{85!50!60! }{ 40!45!}$$

The correct answer is option A.
Thus, option A is the correct answer.

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Joined: 15 Nov 2019
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Re: A double decked bus can accommodate 110 passengers, 50 in the upper de  [#permalink]

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24 Nov 2019, 04:47
Maybe someone can explain to me why my approach seems to be wrong:

1. If 15 refuse to sit in the upper deck, we have to calculate in how many ways we can select 50 people out of 110-15=95 --> 95! / (50!*45!)

2. If 10 refuse to sit in the lower deck, we have to calculate in how many ways we can select 60 people out of 110-10=100 --> 100! / (60!*40!)

3. This would give me --> ( 95! / (50!*45!) ) * ( 100! / (60!*40!) ) ways
Re: A double decked bus can accommodate 110 passengers, 50 in the upper de   [#permalink] 24 Nov 2019, 04:47
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