To evaluate permutations and combinations, we use the following formulas:
[tex]\begin{gathered} nPr=\frac{n!}{(n-r)!} \\ \\ nCr=\frac{n!}{r!(n-r)!} \end{gathered}[/tex]where n is the total number of objects and r is the number of objects selected from the set.
So, to find 7P4 we have:
[tex]\begin{gathered} 7P4=\frac{7!}{(7-4)!} \\ \\ 7P4=\frac{7!}{3!} \\ \\ 7P4=\frac{7\cdot6\cdot5\cdot4\cdot3!}{3!} \\ \\ 7P4=7\cdot6\cdot5\cdot4 \\ \\ 7P4=840 \end{gathered}[/tex]To solve for 10C3, we have:
[tex]\begin{gathered} 10C3=\frac{10!}{3!(10-3)!} \\ \\ 10C3=\frac{10!}{3!7!} \\ \\ 10C3=\frac{10\cdot9\cdot8\cdot7!}{3\cdot2\cdot1\cdot7!} \\ \\ 10C3=10\cdot3\cdot4 \\ \\ 10C3=120 \end{gathered}[/tex]The answers are 840 and 120.
