A school has 480 girls and 520 boys. how many committees of 5 can u make if at least one boy must be on each committee.
There are two ways to do it:
1) Simpler Way (solving it indirectly):
Find the total number of committees and then subtracts the number of committees which are all girls, as
$\displaystyle \binom{1000}{5}-\binom{480}{5}$
OR
2) Solving it directly:
$\displaystyle \binom{520}{1}\binom{480}{4}+\binom{520}{2}\binom{ 480}{3}+\binom{520}{3}\binom{480}{2}+\binom{520}{4 }\binom{480}{1}+\binom{520}{5}$
Hello, wikji!
Who assigned this problem?
. . In fact, who wrote it?
A school has 480 girls and 520 boys. How many committees of five
can be made if at least one boy must be on each committee?
There are: .$\displaystyle {1000\choose5} \:\approx\:82.5\times 10^{11}$ possible committes.
The opposite of "at least one boy" is "no boys" (all girls).
There are: .$\displaystyle {480\choose5} \;\approx\:2.1\times10^{11}$ all-girl committees.
There are about: .$\displaystyle 8\times 10^{12}$ (eight trillion) committees with at least one boy.