help with combinationss (1 Viewer)

nick90

Member
Joined
Feb 7, 2007
Messages
37
Gender
Male
HSC
2008
Gday everyone, I'm having a few problems with understanding combinations - it's very frustrating! I was wondering if you can help me with understanding some of these problems from the fitzpatrick textbook.

Ex 28 (b)

Q 8: In how many ways can 3 cards be selected from a pack of 52 playing cards if (i) at least one of them is an ace?


I can't understand why in (i) you can't just go 4C1 x 51C2 ? I eventually worked out the correct answer another way, but I don't understand why 4C1 x 51C2 doesn't work.


Q 10: A committee of 7 politicians is chosen from 10 liberal members, 8 labor members and 5 independents. In how many ways can this be done so as to include exactly 1 independent and at least 3 liberal members and at least 1 labor member?


I just have no idea how to approach this one.

Any and all help is appreciated! Thanks.
 

denoz

Member
Joined
Oct 4, 2006
Messages
48
Gender
Male
HSC
2007
For question 10:

(5 x 10C3 x 8C3) + (5 x 10C4 x 8C2) + (5 x 10C5 x 8C1) = 73080
 

alez

feel like an angel
Joined
Mar 26, 2007
Messages
276
Gender
Female
HSC
2008
lol i did those exact questions. anyway

ok question 8 u have to work out whether its 1,2 or 3 aces separately.

1 ace - 4C1 X 48C2 =4512
2 ACE - 4C2 X 48C1 = 288
3 ACE - 4C3 = 4
total = 4804

u cant just go 4C1 cuz u have to work out how many u can. 4C1 is just one set of possibilities. it says at least so u have to work out 1,2 or 3 aces

10 - i draw up tables with all the possibilities its easier and then work it out

ind lib lab
1 3 3
1 4 2
1 5 1

this way its easier to see what u have to work out. 5C1 X 10C3 X 8C3 for the first one and then add all answers up
 

nick90

Member
Joined
Feb 7, 2007
Messages
37
Gender
Male
HSC
2008
thanks a lot for the help with that. I think it's helped my understanding a fair bit.


Cheers
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top