# help with math question - combination

• Jan 23rd 2009, 09:23 AM
tukilala
help with math question - combination
12 men and 10 women should be seated around a round table.
How many ways are there to do that …

a. … if we require each woman to sit between two men?
b. … if we require all women to sit adjacently (i.e. all women, except for two, sit between two women)?
c. … if they are 10 married couples (and two single males) and each couple is to be seated adjacently?

can u explain me please how u got to the answers..... thnx
• Jan 23rd 2009, 09:38 AM
Plato
Quote:

Originally Posted by tukilala
12 men and 10 women should be seated around a round table. How many ways are there to do that …
c. … if they are 10 married couples (and two single males) and each couple is to be seated adjacently?

There are $\left( {N - 1} \right)!$ ways to arrange $N$ distinct items is a circle.
Here you have twelve distinct items: ten couples and two singles.
There are $\left( {11} \right)!$ ways to arrange them so that the husband is on his wife’s right.
But that last condition may hold. So we add a factor $2^{10} \left( {11} \right)!$ to account for the ways each couple may choose to be seated.
• Jan 23rd 2009, 10:58 AM
Soroban
Hello, tukilala!

Quote:

12 men and 10 women should be seated around a round table.
How many ways are there to do that …

a) if we require each woman to sit between two men?

Seat the 12 men around the table.
. . There are $11!$ possible seatings.

Place a chair between each of the men.
Assign the ten women to ten of the twelve empty chairs:
.There are $P(12,10) \,=\,\frac{12!}{2!}$ ways.

Answer: . $\frac{(11!)(12!)}{2!}$

Quote:

b) if we require all women to sit adjacently?

Duct-tape the ten women together.
Then we have 13 "people" to seat around the table.
. . There are $12!$ ways.

But the ten women can be ordered in $10!$ ways.

Answer: . $(12!)(10!)$

Quote:

c. if they are 10 married couples and two single males
and each couple is to be seated adjacently?

Duct-tape the married couples together.
Then we have 12 "people" to seat.
. . There are $11!$ ways.

But each couple has two possible orders: husband-wife or wife-husband.
. . And there are: $2^{10}$ orderings.

Answer: . $(11!)(2^{10})$

• Jan 23rd 2009, 11:19 AM
tukilala
i dont understand what have u done in question a
what is p(12,10) and why its equal to 12!/2! ways
and why is the ansewer is ((11!)(12!))/2!

thnx