Combinations Question Urgently

[snaidu228]

Homework Statement

There are a total of 30 members of Parliament in a new country called JanesWorld. Among these deputies, there are 10 from the “Conservative Janes” , 8 from the “Progressive Janes” and 12 independents . How many ways can Queen Jane form a parliamentary committee of 11 parliamentarians, if:
(a) the committee must contain at least one independent ; (b) the committee must contain at most two independents ; (c) the committee has exactly six Conservative Janes, but the Conservative X and the Progressive Y cannot be on the committee at the same time.

Homework Equations

Combinations and Permutations

The Attempt at a Solution

a) Total number of possible committees= C(30,11)
# of committees without independents= C(18,11)
Therefore, numer of committees with atleast one independent= C(30,11)- C(18,11)

b) Total number of possible commitees= C( 30,11)
Total # of committes without independents= C(18,11)
Total # of comm. with 1 independent= C( 12,1) x C( 18,10) ?
Total number of committees with 2 independents= C( 12,2) x C( 18,9) ?

c) Choosing 6 out of 10 conservatives= C( 10,6)
Total number of comittees= C(30,11)= C(10,6)+ C(20,5)
Possibility of X being in committee= C( 10,1)
Possiblity of Y being in committee= C( 20,8) x C(8,1) ?

 

[LCKurtz]

Homework Statement

There are a total of 30 members of Parliament in a new country called JanesWorld. Among these deputies, there are 10 from the “Conservative Janes” , 8 from the “Progressive Janes” and 12 independents . How many ways can Queen Jane form a parliamentary committee of 11 parliamentarians, if:
(a) the committee must contain at least one independent ; (b) the committee must contain at most two independents ; (c) the committee has exactly six Conservative Janes, but the Conservative X and the Progressive Y cannot be on the committee at the same time.

Homework Equations

Combinations and Permutations

The Attempt at a Solution

a) Total number of possible committees= C(30,11)
# of committees without independents= C(18,11)
Therefore, numer of committees with atleast one independent= C(30,11)- C(18,11)

That looks good.

b) Total number of possible commitees= C( 30,11)

Total # of committes without independents= C(18,11)
Total # of comm. with 1 independent= C( 12,1) x C( 18,10) ?
Total number of committees with 2 independents= C( 12,2) x C( 18,9) ?

Yes, so add those three up...

c) Choosing 6 out of 10 conservatives= C( 10,6)
Total number of comittees= C(30,11)= C(10,6)+ C(20,5)
Possibility of X being in committee= C( 10,1)
Possiblity of Y being in committee= C( 20,8) x C(8,1) ?

I'm not sure I understand what (c) is asking for. Is it saying that if there are any conservatives there can be no progressives? If so, wouldn't it just be C(10,6)*C(12,5)?