Midair Equipment Seating

It's much easier to count the number of ways that there AREN'T two adjacent seats available. If this happens, then each of the 8 rows contains at least one of the 12 passengers. The only way that this could happen is if 4 rows contain one passenger and the remaining 4 rows contain two passengers. There are $\binom{8}{4}$ ways to choose the rows with only one passenger seated, $2^4$ ways to choose which seat those passengers fill in their row, and $12!$ ways to arrange the people in their seats after the fact. Since this is a probability problem, we must also find the number of ways to sit everybody with no restrictions. This is simply $16\cdot 15\cdot \cdots \cdot 5=\dfrac{16!}{4!}$ . We can now find the probability that there aren't two seats adjacent:

$\dfrac{\binom{8}{4}\cdot 2^4\cdot 12!}{\frac{16!}{4!}}=\dfrac{\binom{8}{4}\cdot 16}{\binom{16}{4}}=\dfrac{8\cdot 7\cdot 6\cdot 5\cdot 16}{16\cdot 15\cdot 14\cdot 13}.$

After some common factor cancellation, we are left with the fraction $\frac{8}{13}$ . Thus the probability that there ARE two adjacent seats available is $\frac{5}{13}$ , which gives an answer of 18.

We are looking for the ratio of valid seatings to total seatings. To find the total number of seatings, we multiply the number of ways to choose which seats are reserved for people by the number of ways to assign people to these seats. There are ${16 \choose 12}$ ways to choose 12 seats for the 16 people to sit in. Also, there are $12!$ ways to assign people to sit in those seats. Thus, the total number of ways to seat 12 people in 16 seats is $12!{16 \choose 12}$ .

The number of valid seatings is the number of ways we can have an empty row (here, empty means not containing people). Note that we will have four empty seats, so we can have up to 2 empty rows. If we choose 1 of the 8 rows to be empty, we have 14 remaining seats in which to seat the people, and then we must arrange the people. Thus, it is tempting to say that the total number of valid seatings is $12! \cdot8{14 \choose 12}$ . However, we have overcounted each case in which there are two empty rows. In these cases, we must choose 2 of the 8 rows to be empty, and then we have to arrange the people in the remaining 12 seats. Hence, we must subtract $12!{8 \choose 2}$ from our preliminary count. Therefore, there are $12! \cdot8{14 \choose 12} - 12!{8 \choose 2}$ valid seatings.

Finally, we take the ratio of these numbers and simplify greatly to get the desired probability:

$\frac {12! \cdot8{14 \choose 12} - 12!{8 \choose 2}} {12!{16 \choose 12}} = \frac {8{14 \choose 12} - {8 \choose 2}} {{16 \choose 12}} = \frac {8 \cdot \frac {14\cdot 13} {2} - \frac {8 \cdot 7} {2}} {\frac {16 \cdot 15 \cdot 14 \cdot 13} {4!}} = \frac {700} {1820} = \boxed {\frac {5} {13}}$

We first calculate the no. of arrangements possible when no 2 seats are available for the equipment.

As no seat is available for equipment all rows must have atleast 1 seat occupied with passenger and there must be exactly 4 rows containing 2 occupied seats. We can select these 4 rows in $^8C_4$ ways. Now we are left with the rows containing only 1 occupied seat. This seat can be selected in 2 ways for each row. So, total no. of ways=2x2x2x2=16. We are done with selecting the seats. Now, the passengers can be seated in 12! ways.

Now, we calculate total no. of arrangements which is obviously $(^{16}C_{12})*12!$

So, Probability that there are no two seats in a row available for equipment =(No. of favourable arrangements)/(Total no. of arrangements) = $\frac{8}{13}$

Thus, probability that there are 2 seats available for equipment= $\frac{5}{13}$ =>a+b=18

Case 1: There are 2 rows empty, rest all filled up. There are 8 choose 2 = 28 ways in this case. $$ Case 2: There is a row empty, and 2 rows with an occupied seat. Then there are 8 (7 choose 2) 4 = 672 ways. $$ Thus the probability is (28+672)/(total ways, which is 16 choose 4) = 5/13.

Clearly, there are 16C12=1820 ways to choose spots for our 12 passengers. (We will assume they are indistinguishable.) We proceed using complementary counting. We must count the number of ways the luggage will not fit. This means that there is at least one person in each row of 2. After we place 8 people, one in each row, there are four people left. These people can go in any spots left. Thus, we will have four full rows, and four rows with 1 person. There are 8C4 ways to choose the four rows that are full, and 2^4 way to choose which seat the person alone in his/her row will sit. Thus, there are (8C4)*2^4=1120 ways to sit the people so our luggage will not fit. Thus, the probability that it will not fit is 1120/1820=8/13. Since we used complementary counting, the probability that the luggage DOES fit is 1- 8/13=5/13. Thus, our answer is 5+13=18.

Rather than thinking of allocating passengers to seats, we think of "allocating" 4 empty seats.

First, we calculate the number of ways to allocate 4 empty seats such that at least one row is empty. After allocating an empty row (there are 8 possible rows to allocate to), there are 14 seats left and 2 more empty seats to allocate. This amounts to $8 \times \dbinom{14}{2}=728$ ways.

However, this method would have counted the allocations where two rows are empty twice. The number of such allocations (choosing two rows out of 8) is $\dbinom{8}{2}=28$ . Thus the actual number of allocations with one row empty is $728-28=700$ .

The total number of possible allocations is $\dbinom{16}{4}=1820$ . Thus the probability that the computer allocates one row empty is $\frac{700}{1820} = \frac{5}{13}$ .

The number of ways in which $12$ passenger can sit in the $16$ seats randomly is $\binom {16}{12} \cdot 12!$ (first choosing 12 seats from the $16$ , then arranging the passengers in $12!$ ways)

Now, if we keep a row empty for the equipment and make the passengers sit in the other $7$ rows, they can seat in total $8 \cdot \binom {14}{12} \cdot 12!$ (first choosing a row for the equipment in $8$ ways, then choosing seats for the passengers, and make them sit there).

But, a case is counted twice here. In the arrangements where two of the rows are empty, we have counted them twice, once as the first row empty, and again as the second row empty. So we have to subtract the number once to remove the double counting. Now if two rows are empty, then the total number of arrangement is $\binom {8}{2} \cdot 12!$ (first choosing two rows to set as empty, then filling the other $6$ rows in $12!$ ways with the passnegers).

So, the probability we want, is $\frac{8 \cdot \binom {14}{12} \cdot 12! - \binom {8}{2} \cdot 12!} {12! \cdot \binom {16}{12}}$ = $\frac{8 \cdot \binom {14}{12} - \binom {8}{2} } { \binom {16}{12}}$ = $\frac {5}{13}$

So our desired answer is $5+13=18$

[Note: in this problem, whether the passengers are the identical or different doesn't matter, because the probability only depends on the seats which are free]

There are 16C4 (equals to 1820) way to arrange the empty seats on the plane.

First, we need to imagine the seating arrangements as two columns of eight.

It is easier to work out seating arrangements which are not allowed. I do it in the following procedure:

1) no empty seats on the left column, means all empty seats are on the right column. Number of possible arrangements = 1 x 8C4 = 70

2) 1 empty seat on the left column, 3 empty seats on the right column in any arrangement, as long as not on the exact right of the empty left seat. Number of possible arrangements = 8C1 x 7C3 = 280

3) 2 empty seats on left column, 2 empty on the right Number of possible arrangements = 8C2 x 6C2 = 420

3) 3 empty seats on the left Number of possible arrangements = 8C3 x 5C1 = 280

4) all empty seats on the left Number of possible arrangements = 8C4 = 70

Total number of seat arrangements which made placing the equipment impossible = 1120, which means that there are 700 possible ways that there are 2 seats available on the same row to place the equipment.

The probability is 700/1820, simplifying will result in 5/13, which means that the sum of a and b is 18

Consider the 4 empty seats. Suppose there is space for the equipment. There are 8 possible rows for the equipment. So there are 8 * 14C2 = 728 possible configurations. However the cases where there are two rows of empty seats are double-counted. Hence in fact there are 700 ways. The probability is 700 / (16C4) = 5 / 13 in the simplest form. So a + b = 5 + 13 = 18.

Since there 12 passengers on that flight, the 4 vacant seats can be arranged in C(16,4) ways. However, the challenge is to accommodate a large equipment requiring two seats of the same row. There are only 8 rows, and if there are exactly two pairs of vacant seats (each pair contains two seats of the same row) there will be C(8,2) ways to accommodate the equipment. On the other hand, if there should be only one pair of seats of the same row vacant for that baggage, the other two vacant seats must be arranged such that they will not belong to the same row. Let's say the baggage is situated on the first row, the other two vacant seats can be arranged in C(14,2) ways, and there are 7 instances that these seats will be situated in the same row. Therefore, there are C(14,2) minus 7 ways, and this is the same throughout the remaining seven rows. Thus, there are 8 times [C(14,2)-7] ways that exactly one pair of seat is available for the equipment. Adding all the possible ways, we have C(8,2) + 8[C(14,2)-7], and to get the probability we have to divide the sum by C(16,4). Therefore, as computation goes, a/b = 5/13, thus, a+b=18.

Well if we find the other half(probability of seats being not available) then we can find the probability for seats being available by subtracting it from 1(as there are only two possible cases). So seats will not be available wen every row has one occupied seat. Now there are 12 people and 8 \times 2 seats,so 4 rows will be occupied and other four rows will have one person sitting(in each). We can chose 4 rows in {8 \choose 4} and 4 people can sit in the rows in 2^4,so both can be done in 2^4 \times {8 \choose 4}.

Total possible arrangements of seats {16 \choose 12}

The probability \frac {2^4 \times {8 \choose 4}}{{16 \choose 12}}.

Thus the required probability 1- \frac {2^4 \times {8 \choose 4}}{{16 \choose 12}} = \frac {5}{13}

no. of passengers can seat in plane =16 no. of passengers available=12 so 4 seats will be remained no of ways in which 4 seat will remain vacant = 16c4= 1820

case:1(when 2 rows are vacant) no. of ways in which 2 rows will remain vacant = 8c2 =28

case:2(when only 1 row and any 2 extra seats are vacant) no of ways in which only one seat will remain vacant=8c1 (14c2-7) 8 (91-7)=8*84=672 then by adding both we will get the no. of ways in which at least 1 row will remain vacant =672+28=700

(because first we will select 1 row which is needed for machine then we can select any 2 from remaining 14 but we have to exclude 2 seats in same row because we have included them in 1st case)

so probability=700/1820=5/13 so a=5 and b=13 so a+b =18

in this case we can ignore the permutation of the passengers, because 12! in both numerator and denominator will be eventually cross out each other.

so the n(S) = 16c12 = 1820 now we need to think about the seating RIGHT x x x x x x x x LEFT x x x x o o o o *("x" represents "seated", "o" means empty) this is the only arrangement that no two seats in the same row, A'. n(A') = 8c4 * 2 * 2 * 2 * 2 =1120 [the arrangement of 4 rows of double passengers among the 8 rows * (left or right for the single passenger)^4]

so n(A) = n(S) - n(A') = 1820 - 1120 = 700 P(A) = 700/1820 = 5/13

a+b = 5+13 = 18

Solution 1: There are 16 seats on the plane. Without loss of generality we can choose any seat to be empty. For the second empty seat, there is a $\frac{14}{15}$ chance that it is not in the same row. Assuming the first two are in different rows, for the third empty seat, there is a $\frac{12}{14}$ chance that it is not in the same row as either of the first two. Assuming none of the first three are in the same row, there is a $\frac{10}{13}$ chance the last empty seat is in a different row. Thus, the probability that all empty seats are in different rows is $\frac{14}{15}\cdot \frac{12}{14} \cdot \frac{10}{13} = \frac{8}{13}$ , so the probability some pair is in the same row is $1 -\frac{8}{13} = \frac{5}{13}$ . Hence, $a+b = 5+13 = 18$ .

_Solution 2: _ We will use the principle of inclusion/exclusion to determine this. There are $\binom{16}{12} = 1820$ total ways to choose which seats the passengers are in. There are $\binom{14}{12} = 91$ choices for which seats are filled in which a particular row is empty. For any pair of rows, there is 1 way to make both of those rows empty. We cannot have more than two rows empty at a time. Thus, the principle of inclusion/exclusion tells us that there are $8(91) - 28(1) = 700$ ways in which there will be one empty row. Thus, the probability is $\frac{700}{1820} = \frac{5}{13}$ . Hence, $a+b = 5 + 13 = 18$ .