Probability Problem

miller wrote:A five-person team is being created. The possible team members include 4 from City A, 3 from City B and 3 from City C.

1) How many ways can the team be selected so that at least 1 member from City C is included?

2) How many ways can the team be selected so that at least 1 member from City B and at least 1 member from City C are included?

Can anyone explain how to calculate???

for the frist question.

at least 1 menber from C and others 4 left from both a and b
ok, we dont care of the number of A and B much thus we can solve it by this way : combine A and B together we will have 7 prople from group (A.B) now the maths question is turned into how many ways can the team be selected from c and (a,B)so that at least 1 mention from C .
+ select 1 from C and 4 from (AB)
+select 2 from C and 3 from (AB)
seledct 3 from C and 2 from (AB)

that is what i will solve...

Hi,
My approach is to use elimination technique.

1. To find the number of ways that at least 1 member from City C is included, find the number of ways that no member from City C is included.
Then subtract the no. of total ways to the no. of ways that no member from City C is included, we will get the number of ways that at least 1 member from City C is included.

This is based on the basic rule:
Some groups have C, some groups don't. However, total no. always = no. of have C + no. of don't have C

Now, if there's no memember from City C in the team, there'll be only 4 from City A, 3 from City B left.
No. of ways that there's no memember from City C in the team: 7C5 = 21

Total No. of ways: 10C5 = 252

No. of ways that there's at least one member from city C: 252 - 21 = 231
Or you can solve this in the way that diebeatsthegmat said:
3C1*7C4 = 105
3C2*7C3 = 105
3C3*7C2 = 21
Total: 231


2. Using eliminate:
Eliminated cases: 4A and (B or C), 3A2B, 3A2C, 2A3C, 2A3B
Ways to be eliminated:
4A and (B or C) : 4C4*6C1 = 6
3A2B: 4C3*3C2 = 12
3A2C: 4C3*3C2 = 12
2A3C: 4C2*3C3 = 6
2A3B: 4C2*3C3 = 6
Total ways to be eliminated : 42
So total ways that at least 1 member from City B and at least 1 member from City C are included = 252 - 42 = 210

Using diebeatsthegmat's technique:
1C1B3A : 3C1*3C1*4C3 = 36
1C2B2A: 3C1*3C2*4C2 = 54
2C1B2A: 3C2*3C1*4C2 = 54
xxxx1A: 6C4*4C1 = 60 ( x stands for either B or C. When pick 4 from 3 B and 3 C, clearly that B & C will both be picked)
xxxxx: 6C5= 6
Total: 210