A subway car has 10 individual seats, with 5 in front and 5 in back. From 10 passengers, 4 prefers front seat, 3 prefers back seat and the others have no preference. In how many ways can passengers be seated, respecting preferences?Attempt: I want to solve using only the notion of Fundamental principle of counting:Let F be each of the 4 passengers who want to sit in the front, N the passengers who have no preference, and C the passengers who want to sit back. Now, since we have 32˙ possibilities to take a passenger without preference (it will be multiplied by the final result) we will take one of the passengers N to sit in the front seat, we will have, by the multiplicative principle: 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1Now N − 1 = 2 passengers without preference are left, and they will be exchanged with passengers who want to sit back: 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1But if you go on like that, and at the end of it all, multiplying it by 6 I don't get the right result. What am I wrong?The answer is 43200

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A subway car has 10 individual seats, with 5 in front and 5 in back. From 10 ​passengers, 4 prefers front seat, 3 prefers back seat and the others have no preference. In how many ways can passengers be seated, respecting preferences?Attempt: I want to solve using only the notion of Fundamental principle of counting:Let   F be each of the 4 passengers who want to sit in the front,   N the passengers who have no preference, and   C the passengers who want to sit back. Now, since we have 32˙ possibilities to take a passenger without preference (it will be multiplied by the final result) we will take one of the passengers N to sit in the front seat, we will have, by the multiplicative principle:  5  ⋅  4  ⋅  3  ⋅  2  ⋅  1Now   N  −  1  =  2 passengers without preference are left, and they will be exchanged with passengers who want to sit back:  5  ⋅  4  ⋅  3  ⋅  2  ⋅  1But if you go on like that, and at the end of it all, multiplying it by 6 I don&#039;t get the right result. What am I wrong?The answer is 43200

lodosr · Accepted Answer

Front row seating - Once you have chosen a person with no preference to seat along with 4 persons who prefer front row, there are   5  ! ways to seat them as you counted.Back row seating - for remaining 5 including 3 who prefer to sit in the back row, there is again   5  ! ways to seat them as you counted correctly.The only part that you missed is to choose one of the three persons with no preference for the front row.So the answer should be      3  ⋅  5  !  ⋅  5  !  =  43200Another approach for counting can be to seat 4 people with front row preference and 3 people with back row preference. Then seat rest 3 folks in remaining seats.                                          (                                    5          4                                      )                                ⋅    4    !    ⋅                                        (                                    5          3                                      )                                ⋅    3    !    ⋅    3    !    =    43200

excluderho · Accepted Answer

Isn’t it just   P  (  5  ,  4  )  ×  P  (  5  ,  3  )  ×  (  10  −  4  −  3  )  ! ? where   P means permutations.The method in the OP just confuses me and I don’t understand what the OP is trying.

A subway car has 10 individual seats, with 5 in front and 5 in back. From 10 passengers, 4 prefers

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