Permutations and Combinations

[w.tmendis]

Eight persons, consisting of four married couples, are to be seated in a row of eight chairs. How many seating arrangements are there if:

i. There are no other restrictions
ii. The men must sit together and the women must sit together
iii. men must sit together
iv. The spouses in each married couple must sit together

I mess with this problem. Please help me. Have a nice day.

Thilini

[dw6586]

can you please clarify the difference between ii and iii? doesn't the men must sit together also mean the same as the men must sit together and the women must sit together? and does the order, or arrangement, of the seating matter in any of the instances?

[garima_aries01]

Eight persons, consisting of four married couples, are to be seated in a row of eight chairs. How many seating arrangements are there if:

i. There are no other restrictions
ii. The men must sit together and the women must sit together
iii. men must sit together
iv. The spouses in each married couple must sit together

I mess with this problem. Please help me. Have a nice day.

Thilini

===============================
There are 8 seats in a row which can be represented as

- - - - - - - -(i) no restrictions: 8![1st can be occupied in 8 ways , 2nd seat can be occupied in 7 ways and so on]

(ii) The men must sit together and the women must sit together: 2x4!x4!
[ - - - - - - - -]
- - - - - - - -
men can occupy blue seats and women can occupy red seats in 2 ways
men can occupy blue seats in 4! ways
women can occupy red seats in 4! ways

so 2x4!x4!

iii) men must sit together : 5x4!x4!=> men must be together , women may or may not be together

color=#400080] _ _ _ _[/color] _ _ _ _
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _men can occupy blue seats in 4! ways
women can occupy left over seats in 4! ways
blue seats can be in 5 ways
so 5x4!x4!

iv)The spouses in each married couple must sit together
:16x4!
imagine u have tied a rope to each couple
so in total there are 4 places

_ _ _ _ _ _ _ _

these couples can be arranged in 4! ways
each couple in rope can be further placed in 2 ways _ _
_ _
and there are 4 couples

2x2x2x2x4! = 16x4!

[jnelson0612]

Garima, very nice! Thanks!