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

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Bunuel
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A double decked bus can accommodate 110 passengers, 50 in the upper de [#permalink] New post   12 Nov 2019, 02:40
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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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A double decked bus can accommodate 110 passengers, 50 in the upper de [#permalink] New post   17 Jan 2020, 07:05
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LevanKhukhunashvili wrote:
Sir chetan2u please introduce light in my darkened mind

My reasoning was as follows:
If we arrange upper deck the lower one would be arranged automatically and vice versa
We have got 15 people in the upper deck and 10 in lower, so they have seated there and we cant move them by force
Now we are left to arrange 85 people
There is no point which deck we arrange

Upper deck: arranging 85 people on 40 seats: 85!/40!45!
Lower deck: arrangind 85 people on 45 seats: 85!/45!40!

IMO Ans: B

I was concerned after reading EgmatQuantExpert's replay


Hi,

Actually answer depends on what has been asked..

If ways means combinations then \(\frac{85!}{40!45!}\) is correct.
You have shifted 15 and 10 at respective decks and can put remaining 85 in \(\frac{85!}{40!45!}\) ways.
So what we have is 50 and 60 in different decks

But say we are looking for the ways to arrange them that is order matters. then
After distributing them in groups of 50 and 60, these 50 and 60 can be arranged in 50! and 60! ways and then answer becomes \(\frac{85!}{40!45!}*50!*60!\)
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Re: A double decked bus can accommodate 110 passengers, 50 in the upper de [#permalink] New post   12 Nov 2019, 02:54
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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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Re: A double decked bus can accommodate 110 passengers, 50 in the upper de [#permalink] New post   13 Nov 2019, 03:46
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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.

Correct Answer: Option A
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Re: A double decked bus can accommodate 110 passengers, 50 in the upper de [#permalink] New post   24 Nov 2019, 04:47
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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
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Re: A double decked bus can accommodate 110 passengers, 50 in the upper de [#permalink] New post   06 Jan 2020, 03:37
85C45 * 45C45 ------> B
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A double decked bus can accommodate 110 passengers, 50 in the upper de [#permalink] New post   06 Jan 2020, 04:59
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15 refuse to be in the upper deck -> Our only choice is to accommodate these 15 in the lower deck

10 refuse to be in the lower deck -> Our only choice is to accommodate these 10 in the upper deck

Since the decks for these 15+10=25 people are already fixed, we are left with 110-25=85 people to accommodate.

Since 10 have already been accommodated in the upper deck, we have 50-10=40 more places to fill in the upper deck and all of the remaining 45 will be in the lower deck.

Number of ways to pick 40 people from a total of 85 = 85C45 = 85!/45!40!

Answer is (B)
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Re: A double decked bus can accommodate 110 passengers, 50 in the upper de [#permalink] New post   17 Jan 2020, 04:05
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Sir chetan2u please introduce light in my darkened mind

My reasoning was as follows:
If we arrange upper deck the lower one would be arranged automatically and vice versa
We have got 15 people in the upper deck and 10 in lower, so they have seated there and we cant move them by force
Now we are left to arrange 85 people
There is no point which deck we arrange

Upper deck: arranging 85 people on 40 seats: 85!/40!45!
Lower deck: arrangind 85 people on 45 seats: 85!/45!40!

IMO Ans: B

I was concerned after reading EgmatQuantExpert's replay
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Re: A double decked bus can accommodate 110 passengers, 50 in the upper de [#permalink] New post   04 Oct 2023, 00:08
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Re: A double decked bus can accommodate 110 passengers, 50 in the upper de [#permalink]
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