# Thread: More Data .. Can you figure it out?

1. ## More Data .. Can you figure it out?

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.

2. 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:
$\displaystyle \binom{520}{1}\binom{480}{4}+\binom{520}{2}\binom{ 480}{3}+\binom{520}{3}\binom{480}{2}+\binom{520}{4 }\binom{480}{1}+\binom{520}{5}$

3. 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.

4. Originally Posted by Soroban
Hello, wikji!

Who assigned this problem?
. . In fact, who wrote it?

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.

How do u get this ... Sry but im the worst math student of all time ... here are about: .8\times 10^{12} (eight trillion) committees with at least one boy.