Help with some harder perms and combs??? (1 Viewer)

[pauloz!]

Hey i dont have the answers to these 2 perms and combs questions so im not sure whether i am right or wrong. Could you guys try these out and show me your answers just so i can confirm whether i am doing these questions right??

1. 8 people are to be seated around a table. If the 8 people are 4 married couples, how many ways can they be seated if no husband and wife, as well as no 2 men are to be in adjacent seats?

2. 12 girls and 15 boys attend a school party. In how many ways can 4 pairs be selected to dance? (each pair being 2 people of the opposite sex)

Thanks

 

[badquinton304]

Draw up a circle with M1, M2, M3, M4, F1, F2, F3, F4.
M1 next to all F except F1
M2 next to all F except F2
M3 next to all F except F3
M4 next to all F except F4
F1 not next to M1
F2 not next to M2
F3 not next to M3
F4 not next to M4

You also have to remember that Since no to males can sit together, no 2 females will be able to sit together (because that will mean some males will sit together)

So there is 12 different ways for males (3 each) to sit and 3 different ways for each female so 12. But since the same sex may not be seated next to each other the male female combinations overlap leading to 12. I hope thats correct it think I could have went wrong somewhere so take it with a grain of salt but thats just my answer.

For the second one I think you just would find the combinations of picking 2 girls out of 12 (66 combinations) and then multiply by the number of combinations picking 2 out of 15 boys (105 combinations) so I got 6930.
Again my answer may not be correct I haven't started 4 unit yet I just wanted some fun questions to do.

 

[kony]

q2.

first let's pick the 8 people that are to dance.
we have 12C4 girls x 15C4 boys = 675675 ways of choosing the 8 people.


within the 8 people, we want to find the number of ways to couple them.
i had to think about this for a while - i think the following approach is correct.

imagine if we place a boy in each corner of a room. where they stand doesn't matter. now consider the numer of ways for the girls to stand such that they go join a boy at a corner.

this is clearly 4P4 as order DOES matter here.

therefore, there are 675675 x 24 = 16216200 ways.

 

[kony]

q1.

a bit of a strange question i think: this is my approach.

first it is obvious that the seating must be that the men and women alternate.

consider the males seated first, with Male 1 sitting at a particular spot.

.......M1

M............M

...... M

there are 3P3 ways to arrange the men around a circle.

now let's put a random one of these choices down:

.......M1

M4..........M2

...... M3

consider the space between M1 and M2. we have 2 choices here, either F3 or F4. let's suppose we put down F4.

now consider the space between M2 and M3. we also have 2 choices, either F1 or F4. but F4 has already been taken. therefore, we have to choice F1.

consider space between M3 and M4. it's either F1 or F2. but F1 was chosen already. so we have to choose F2. the last woman has no choice. so F3 sits between M1 and M4.

what the last 3 paragraph showed was that: as soon as you sit down one woman, the rest of the seating is predetermined. this means for each male seating plan, there are only 2 ways to sit down the women.

hence, 3P3 x 2 = 12.

badquinton also got 12, but i don't understand what he was talking about.

 

[badquinton304]

q1.

a bit of a strange question i think: this is my approach.

first it is obvious that the seating must be that the men and women alternate.

consider the males seated first, with Male 1 sitting at a particular spot.

.......M1

M............M

...... M

there are 3P3 ways to arrange the men around a circle.

now let's put a random one of these choices down:

.......M1

M4..........M2

...... M3

consider the space between M1 and M2. we have 2 choices here, either F3 or F4. let's suppose we put down F4.

now consider the space between M2 and M3. we also have 2 choices, either F1 or F4. but F4 has already been taken. therefore, we have to choice F1.

consider space between M3 and M4. it's either F1 or F2. but F1 was chosen already. so we have to choose F2. the last woman has no choice. so F3 sits between M1 and M4.

what the last 3 paragraph showed was that: as soon as you sit down one woman, the rest of the seating is predetermined. this means for each male seating plan, there is only 2 ways to sit down the women.

hence, 3P3 x 2 = 12.

badquinton also got 12, but i don't understand what he was talking about.

Your way is much better mine is way too dodgy.

 

[pauloz!]

THANK YOU SO MUCH!!! Makes alot of sense...i appreciate your help!