Question

Evaluate 7p4 and 10c3.

Answer 1

To evaluate permutations and combinations, we use the following formulas:

`\begin{gathered} nPr=(n!)/((n-r)!) \\ \\ nCr=(n!)/(r!(n-r)!) \end{gathered}`

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:

`\begin{gathered} 7P4=(7!)/((7-4)!) \\ \\ 7P4=(7!)/(3!) \\ \\ 7P4=(7\cdot6\cdot5\cdot4\cdot3!)/(3!) \\ \\ 7P4=7\cdot6\cdot5\cdot4 \\ \\ 7P4=840 \end{gathered}`

To solve for 10C3, we have:

`\begin{gathered} 10C3=(10!)/(3!(10-3)!) \\ \\ 10C3=(10!)/(3!7!) \\ \\ 10C3=(10\cdot9\cdot8\cdot7!)/(3\cdot2\cdot1\cdot7!) \\ \\ 10C3=10\cdot3\cdot4 \\ \\ 10C3=120 \end{gathered}`

The answers are 840 and 120.