A man has 9 frinds : 4 boys and 5 girls. In how many ways can he invite them, if there have to be exactly 3 girls in the invitees?
My approach -
What I assume that you will give invite to 3 unique girls simultaneously --
so, you need to select 3 girls out of 5 girls -
this is nothing but 5C3 10 ways. Now, 6 are left out. These can be arranged in the 6C6 ways thus -- 6 ways. So, the answer should be 60 ways. However, I don't see any option like 60 for this answer. What am I missing here.
Step 1: Choose exactly $3$ girls from $5: {5 \choose 3} = 10$ ways.
Step 2: Now (since the only necessary condition is satisfied), for each of the $4$ boys, you can choose to invite him or not (2 ways each). Therefore, total ways $ = 2^4 = 16$.
Total ways = $16\times 10 = 160 $ ways.