Question

pleeeease help!
there are 7 teams (A,B,C,D,E,F,G) participating in a swimming tournament; each team 4 members.
Question: 10 members are to be randomly selected from the 28 members. In now many ways can the selection process be conducted if there is a ruling that at most 2 members from the same team may be selected?

Answer 1

actually i'm 100% i did something wrong :/

Answer 2

is it 161

Answer 3

If you were to pick 1 person from each team ... you would have to pick a 2nd person from 3 of the teams

Answer 4

Now,
you could choose 2 people each from five of the teams:

(7 choose 5) ways to pick 5 teams

in 1 team (4choose2) ways to pic 2 people
(4 choose 2)^5 ways to pick 2 people each from the 5 chosen teams

So, number of ways you could choose: 2 people each, from 5 of the 7 teams\[\Large={7\choose5}\times{4\choose2}^5\]\[=21\times6^5\]

Answer 5

OR, you could choose:
..... 2 people each from 4 of the teams, and
..... 1 person each from 2 of the remaining teams.

\[\Large = {7 \choose 4}{4\choose2}^4\times {3\choose 2}{4\choose1}^2\]\[=35\cdot6^4\times3\cdot4^2\]

Answer 6

OR, you could choose:
..... 2 people each from 3 of the teams, and
..... 1 person each from 4(all) of the remaining teams.

\[\Large ={7\choose3}{4\choose2}^3\times \cancel{4\choose4} {4\choose1}^4\]\[=35\cdot6^3\times 1\cdot 4^4\]

Answer 7

add them all up ... and you have you answer!!!
\[\Huge \color{purple} ✿\]
http://www.wolframalpha.com/input/?i=%287choose5%29%284choose2%29%5E5%2B%287choose4%29%284choose2%29%5E4%283choose2%29%284choose1%29%5E2%2B%287choose3%29%284choose2%29%5E3%284choose1%29%5E4
or try http://lmgtfy.com/?q=21*6%5E5+%2B+35*6%5E4*3*4%5E2+%2B+35*6%5E3*4%5E4
\[\Huge ✌☺✌\]

Answer 8

The answer involves the hypergeometric distribution, but let's try to
find it using recursion.

How can you construct a solution for 7 teams and 10 members?:

- Any solution for 6 teams and 10 members still works.

- If you have a solution for 6 teams and 9 members, you can choose 1
member from team 7 (there are four ways to do this).

- If you have a solution for 6 teams and 8 members, you can choose 2
members from team 7 (there are C[4,2] or 6 ways to do this).

In other words,

f[7 teams, 10 members] =
f[6 teams, 10 members] + 4*f[6 teams, 9 members] + 6*f[6 teams, 8 members]

More generically:

f[t,m] = f[t-1,m] + 4*f[t-1,m-1] + 6*f[t-1,m-2]

Our base case is 1 team. We can choose 1 member 4 ways and 2 members 6
ways. We can't choose any teams with 3 or 4 members, since this would
break the rules. And, of course, we can't choose more than 4 members
from a 4 member team.

f[1,1] = 4
f[1,2] = 6
f[1,3] = 0
f[1,4] = 0
f[1,x] = 0 for x>=5

The answer I got after slogging through all the recursion is 3,918,240.

There is almost definitely a better way to do this.

Answer 9

@barrycarter
... i think maybe you forgot to include

f[1,0] = 1
f[2,0] = 1

I really think my answer of 4,275,936 is correct

Answer 10

You're right, I'm wrong. Rats! Adding those two base cases gives me your answer.