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Jaf

Jaf

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For the third one, first find the number of ways Mrs. Brown can sit in the front row. There are only 3 (3P1) possible ways. Now you've assigned a place to her.

Next, since Mrs. Lin needs to sit behind a student, she can take all places except the three seats in the first row and one seat in the second row because Mrs. Brown is already sitting in front of it. There are 10 (10P1) possible ways to seat Mrs. Lin. Now the student sitting in front of Mrs. Lin can be any of the 5 (5C1) students. So one of the students AND Mrs. Lin can be seated in 10P1 x 5C1 ways.
When I first did this question, I went wrong in this part. What I did was I found the number of ways to seat Mrs. Lin by 10P1 and multiplied that answer by 5C1 x 10. (10 because the student has 10 possible ways to be seated assuming both ladies have already sit in a place). This gave me the wrong answer. Why we can't do this is because the student HAS to be in front on Mrs. Lin. So we basically count them both as one group.

Finally, now you have 9 passengers left and 11 unoccupied seats left to seat them. You can seat them in 11P9 ways.
So the answer is (3P1 x 10P1 x 5C1 x 11P9)/(14P12) = 0.0687

It's a bit complicated but I hope you understood.
 
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hard2face

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Jaf said:
For the third one, first find the number of ways Mrs. Brown can sit in the front row. There are only 3 (3P1) possible ways. Now you've assigned a place to her.

Next, since Mrs. Lin needs to sit behind a student, she can take all places except the three seats in the first row and one seat in the second row because Mrs. Brown is already sitting in front of it. There are 10 (10P1) possible ways to seat Mrs. Lin. Now the student sitting in front of Mrs. Lin can be any of the 5 (5C1) students. So one of the students AND Mrs. Lin can be seated in 10P1 x 5C1 ways.
When I first did this question, I went wrong in this part. What I did was I found the number of ways to seat Mrs. Lin by 10P1 and multiplied that answer by 5C1 x 10. (10 because the student has 10 possible ways to be seated assuming both ladies have already sit in a place). This gave me the wrong answer. Why we can't do this is because the student HAS to be in front on Mrs. Lin. So we basically count them both as one group.

Finally, now you have 9 passengers left and 11 unoccupied seats left to seat them. You can seat them in 11P9 ways.
So the answer is (3P1 x 10P1 x 5C1 x 11P9)/(14P12) = 0.0687

It's a bit complicated but I hope you understood.
Click to expand...
 
H

hard2face

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Jaf said:
For the third one, first find the number of ways Mrs. Brown can sit in the front row. There are only 3 (3P1) possible ways. Now you've assigned a place to her.

Next, since Mrs. Lin needs to sit behind a student, she can take all places except the three seats in the first row and one seat in the second row because Mrs. Brown is already sitting in front of it. There are 10 (10P1) possible ways to seat Mrs. Lin. Now the student sitting in front of Mrs. Lin can be any of the 5 (5C1) students. So one of the students AND Mrs. Lin can be seated in 10P1 x 5C1 ways.
When I first did this question, I went wrong in this part. What I did was I found the number of ways to seat Mrs. Lin by 10P1 and multiplied that answer by 5C1 x 10. (10 because the student has 10 possible ways to be seated assuming both ladies have already sit in a place). This gave me the wrong answer. Why we can't do this is because the student HAS to be in front on Mrs. Lin. So we basically count them both as one group.

Finally, now you have 9 passengers left and 11 unoccupied seats left to seat them. You can seat them in 11P9 ways.
So the answer is (3P1 x 10P1 x 5C1 x 11P9)/(14P12) = 0.0687

It's a bit complicated but I hope you understood.
Click to expand...
plz explain me part(ii) of the same question
 
H

hardik

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in part 2
3 business people are sitting in first row so its calculation is 3p3 or 3!
mr and mrs lin , mr and mrs brown sits 2gether(couple) so dey have in total 3 (pair) of seats to seat so 3P2 ways to seat . (3p2 *2*2)because couple might interchange their places.
total 5 students r der and dey tke 5 places. and can interchange their places so 5!

3!*3p2*2*2*5! =17280
 
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