1. ## combinations and permutations

a committe of three people are to be chosen from four married couples. how may way can this committe can be chosen if.

(a) all are equally eligible
(b) the committe must consist of 1 woman and two men
(c) all are equally eligible except that a husband and wife cannot serve on the same committe

for (a) number of ways =3x2x4 = 24

(b) $= \frac{n!}{(n-r)!}$

$=\frac{12!}{(12-3)!}$

$=\frac{12!}{(9)!}$

are these correct .

how would i determine (c)

2. Originally Posted by sigma1
a committe of three people are to be chosen from four married couples. how may way can this committe can be chosen if.

(a) all are equally eligible
(b) the committe must consist of 1 woman and two men
(c) all are equally eligible except that a husband and wife cannot serve on the same committe

for (a) number of ways =3x2x4 = 24

(b) $= \frac{n!}{(n-r)!}$

$=\frac{12!}{(12-3)!}$

$=\frac{12!}{(9)!}$

are these correct .

how would i determine (c)
hi

Assume that no man can be a husband to two wives and vice versa (obviously ) ,

(a) 8C3=56

(b) 4C1 x 4C2 = 24

(c) If a couple is allowed to serve in the committe , the committe will consist of one couple and another person .

pick one couple from 4 , 6 ppl are left to be picked for the remaining one so 4C1 x 6C1=24

If a couple is NOT allowed to serve , there would be 8C3-24=32 ways to form the committe .

3. i am having a bit if difficulty understanding you soloution and how you arrived at your answer. could you please Explain it a bit more i dont understand the use of the Variable C throughout your working,, thanks.

4. Originally Posted by sigma1
i am having a bit if difficulty understanding you soloution and how you arrived at your answer. could you please Explain it a bit more i dont understand the use of the Variable C throughout your working,, thanks.
its a symbol for combination whereby its the same as $n\choose r$

its a symbol for combination whereby its the same as $n\choose r$
6. $\binom{n}{r}=\frac{n!}{r!(n-r)!}$