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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total seats = 110
UP= 50
LD= 60
10 refuse LD and 15 refuse UP
60P15 *50P10* 85!
IMO A
\(\frac{85!50!60!}{40!45!}\)

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

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)