SOLUTION: A group consists of 4 girls and seven boys. In how many ways can a team of 5 members be selected if the team has (i) no girl (ii) at least one boy and one girl. (iii) At least 3 g

That means 'all boys'.  7 boys choose 5 = 7C5 = 21 ways

It's impossible not to have at least 1 boy.
So this is the number of ways to have any 5 of the 11 people
MINUS the cases where they're all boys, which is the result
of part (i)

So it's 

(11 people choose 5) MINUS (7 boys choose 5) = 11C5 - 7C5 = 462-21 = 441 

[Here is a case where you learn that AND implies MULTIPLICATION
and OR implies ADDITION.]

Case 1:  3 girls AND 2 boys

         4 girls choose 3 AND 7 boys choose 2 = (4C3)(7C2) = (4)(21) = 84
             [Notice that we MULTIPLIED due to AND]
OR

Case 2:  4 girls AND 1 boy

         4 girls choose all 4 AND 7 boys choose 1 = 4C4+7C1 = (1)(7) = 7
             [Notice that we MULTIPLIED due to AND]
 
Case 1 OR case 2 = 84+7 = 91
             [Notice that we ADDED due to OR]


Edwin