Originally Posted by

**Soroban** Hello, MathBlaster47!

The first two are Combination problems, not Permutations.

There are: .$\displaystyle _{10}C_3 \:=\:{10\choose3} \:=\:\frac{10!}{3!\,7!} \:=\:120$ ways to choose 3 women.

There are: .$\displaystyle _8C_3 \:=\:{8\choose3} \:=\:\frac{8!}{3!\,5!} \:=\:56$ ways to choose 3 men.

Therefore, there are: .$\displaystyle 120\cdot56 \:=\:6720$ committees with 3 women and 3 men.

There are: .$\displaystyle _{16}C_2 \:=\:{16\choose2} \:=\:\frac{16!}{2!\,14!} \:=\:120 $ possible two-marbles samples.

There are: .$\displaystyle _6C_2 \:=\:{6\choose2} \:=\:\frac{6!}{2!\,4!} \:=\: 15$ ways to get two white marbles.

Therefore: .$\displaystyle P(\text{two whites}) \;=\;\frac{15}{120} \;=\;\frac{1}{8}$

It depends on *which* two rows are interchanged.

If the two rows are an *odd* number of rows apart,

. . the sign of the determinant is changed.

If the two rows are an *even* number of rows apart,

. . the value of the determinant is unchanged.