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Permutations and combinations! Help needed!

Can sumbuddy plz help me with Q6 part ii and iii of the attached paper?? I have attached the marking scheme as well- even though i have the solution, i am still not able to understand how they are doing it... Can somebody plz guide me on that? thanks in advance! =)

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Re: Permutations and combinations! Help needed!

*Q: A small aeroplane has 14 seats for passengers. The seats are arranged in 4 rows of 3 seats and a back*

row of 2 seats (see diagram). 12 passengers board the aeroplane.

(i) How many possible seating arrangements are there for the 12 passengers? Give your answer

correct to 3 signiﬁcant ﬁgures. [2]

These 12 passengers consist of 2 married couples (Mr and Mrs Lin and Mr and Mrs Brown), 5 students

and 3 business people.

(ii) The 3 business people sit in the front row. The 5 students each sit at a window seat. Mr and Mrs

Lin sit in the same row on the same side of the aisle. Mr and Mrs Brown sit in another row on

the same side of the aisle. How many possible seating arrangements are there? [4]

(iii) If, instead, the 12 passengers are seated randomly, ﬁnd the probability that Mrs Lin sits directly

behind a student and Mrs Brown sits in the front row. [4]

Sol:

(ii)

With respect to the (Figure-1) attached.

The three businessmen can choose from the first row (indicated in blue) ways they can be arranged = $\displaystyle 3!$

The married couple can choose from the paired seat (indicated in green) the first pair getting to choose has 3 choice the second pair has 2 choices and the pair can be considered as Mr, Mrs or Mrs, Mr so total arrangements = ($\displaystyle 3$x$\displaystyle 2$)x($\displaystyle 2$x$\displaystyle 2$)

With respect to the (Figure-2)

Now the students get to choose among the remaining 5 window seats (one from the dotted red box, 4 from the solid red box) now the number of ways they can be arranged = $\displaystyle 5!$

$\displaystyle \therefore$ Total no.of arrangements = $\displaystyle 3!$x($\displaystyle 3$x$\displaystyle 2$)x($\displaystyle 2$x$\displaystyle 2$)x$\displaystyle 5!$

(iii) Can be solved on similar line of grouping the cases.

~Kalyan.

Re: Permutations and combinations! Help needed!

Can u plz solve the last part too? :/

Re: Permutations and combinations! Help needed!

Show me your attempt and the difficulty your are facing. I will surely help you.

Kalyan.

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Re: Permutations and combinations! Help needed!

Umm ohk. I somehow sorted that out my self.

Mrs Brown's probability of sitting in the front seat is 3/10. Out of the remaining 13 seats (1 front seat is occupied by Mrs Brown)- there are 10 such seats in which Mrs Lin can place herself so that she will be directly behind a student, so her probability of sitting behind a student is 10/13 and there are 5 students so their probability is 5/12. (Sorry there's a stupid question i have in mind- is this 12 in the denominator is the total number of passengers i.e. 12 or the remaining seats after 2 are occupied?)

So P= 3/10 * 10/13* 5/12= .0687

Using another method, Mrs Brown can sit in 3 front seats, Mrs Lin in 10, and 5 students will obviously occupy 5 seats, the rest of 9 passengers can take 11 seats.

So P= (3*10*5*11P9)/ 14P12

Is what i am understanding out of the solution correct? or there is still a problem? I really hate permutations and combinations- seriously.

P.S. Sorry, i attached the wrong marking scheme before. The correct one is attached with this post.

Re: Permutations and combinations! Help needed!

Quote:

Originally Posted by

**IceDancer91** Mrs Brown's probability of sitting in the front seat is 3/10. ..... So P= 3/10 * 10/13* 5/12= .0687.

This is incorrect the probability of Mrs. Brown is $\displaystyle \frac{3}{14}$ hence over all probability = $\displaystyle \frac{3}{14}.\frac{10}{13}.\frac{5}{12}$

Quote:

Originally Posted by

**IceDancer91** Sorry there's a stupid question i have in mind- is this 12 in the denominator is the total number of passengers i.e. 12 or the remaining seats after 2 are occupied?

The probability is with respect to the seat selection after two are occupied.

Quote:

Originally Posted by

**IceDancer91** 5 students will obviously occupy 5 seats

This argument is incorrect. The correct argument goes like this you have picked $\displaystyle 3$ suitable places for Mrs. Brown and $\displaystyle 10$ suitable places for Mrs.Lin. Now your job is to place one student in front of Mrs.Lin; and the choices you have are $\displaystyle \binom{5}{1}=5$ and the remaining people can pick the seats in $\displaystyle {^{11}P_9}$ ways. And as you have mentioned total possible arrangements are $\displaystyle {^{14}P_{12}}$. Probability of the event is $\displaystyle \frac{3.10.5.{^{11}P_9}}{{^{14}P_{12}}}$

Kalyan.

Re: Permutations and combinations! Help needed!

OH OHK. Thanks a lot for your help. =)

Re: Permutations and combinations! Help needed!

How are there 10 suitable places for Mrs. Lin? Can you elaborate please.

Re: Permutations and combinations! Help needed!

Quote:

Originally Posted by

**kalyanram** This is incorrect the probability of Mrs. Brown is $\displaystyle \frac{3}{14}$ hence over all probability = $\displaystyle \frac{3}{14}.\frac{10}{13}.\frac{5}{12}$

The probability is with respect to the seat selection after two are occupied.

This argument is incorrect. The correct argument goes like this you have picked $\displaystyle 3$ suitable places for Mrs. Brown and $\displaystyle 10$ suitable places for Mrs.Lin. Now your job is to place one student in front of Mrs.Lin; and the choices you have are $\displaystyle \binom{5}{1}=5$ and the remaining people can pick the seats in $\displaystyle {^{11}P_9}$ ways. And as you have mentioned total possible arrangements are $\displaystyle {^{14}P_{12}}$. Probability of the event is $\displaystyle \frac{3.10.5.{^{11}P_9}}{{^{14}P_{12}}}$

Kalyan.

How are there 10 suitable places for Mrs. Lin? Can you elaborate please.

Re: Permutations and combinations! Help needed!

Quote:

Originally Posted by

**ones** How are there 10 suitable places for Mrs. Lin? Can you elaborate please.

Not sure why you are resurrecting a 5 year old post. It may be time to start a new thread.

Look at the diagram that kalyanram posted. One of the front seats is taken by Mrs. Brown, so Mrs. Lin cannot sit behind her. Any other seat where there is an empty seat ahead of her is available (keep in mind, at this point in the selection process, only Mrs. Brown is seated). How many seats have an empty seat in front of them? 11. How many are not behind Mrs. Brown? 10.