Recall that ! means facotrial in math. This means we multiply the number by all positive integers less than itself. In other words, this...

\(\displaystyle \frac{12!}{10!-4!}\)

Becomes

\(\displaystyle \frac{1\cdot 2\cdot3\cdot 4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12}{(1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10)-(1\cdot2\cdot3\cdot 4)}\)

This is a great job for a calculator, which yields:

\(\displaystyle 132.0009\)

Step 1: We need to define factorial (!). The operation (!) means that you multiply the number next to the ! by the next consecutive number down to 1.

Step 2: Expand the equation:

\(\displaystyle 5!=5\cdot4\cdot3\cdot2\cdot1=120\)
\(\displaystyle 3(4!)=\)\(\displaystyle 3(4\cdot 3\cdot 2\cdot 1)=3(24)=72\)

Step 3: Evaluate the expression:

\(\displaystyle 120-72=48\)

The answer to the expression is \(\displaystyle 48\).

\(\displaystyle n!\) is the product of consecutive numbers 1 through \(\displaystyle n.\)

\(\displaystyle \frac{5\times 4\times 3\times 2\times 1}{(3\times 2\times 1)-(2\times 1)} =\)

\(\displaystyle \frac{120}{6-2} = \frac{120}{4} = 30\)

\(\displaystyle n!\) is the product of consecutive numbers 1 through \(\displaystyle n.\)

\(\displaystyle (6\times 5\times 4\times 3\times 2\times 1) - 2(3\times 2\times 1) =\)

\(\displaystyle 720 - 2(6) = 720 -12 = 708\)

Step 1: Rewrite the equation by using rule of factorials...

Rule of factorials: Multiply the number in front of factorial and each number one lower than the rest until I hit \(\displaystyle 1\).

\(\displaystyle \frac {13!}{10!}=\frac {13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}\)

Step 2: Reduce any common numbers (colored in blue):

\(\displaystyle \frac {13\cdot 12\cdot 11\cdot {\color{Blue} 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}{{\color{Blue} 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}\)

We get \(\displaystyle 13\cdot12\cdot11\)!

Step 3: Evaluate what's left

\(\displaystyle 13\cdot12\cdot11=1716\)

The value of the fraction is \(\displaystyle 1716\).

\(\displaystyle _nP_k\) depicts the permutation of n total objects with k objects being arranged.

Thus,

\(\displaystyle _{8}P_{4} = \frac{n!}{k!}=\frac{8!}{4!}\)

The upper factorial is the upper index, and the lower factorial is the difference of the indices. 

\(\displaystyle \frac{8!}{4!} = \frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1}\) 

The \(\displaystyle 4!\) or \(\displaystyle 4\times3\times2\times1\) will cancel out.

\(\displaystyle 8\times7\times6\times5 = 56\times30 = 1,680\)

This particular question is asking one to find the combination. This is because the order of the 1st, 2nd, and 3rd place finishers is not important.

Thus evaluate \(\displaystyle _{12}C_{3}\).

\(\displaystyle _{12}C_{3} = \frac{12!}{(12-3)! 3!}\) 

Because \(\displaystyle 12!\) and \(\displaystyle 9!\) both consist of \(\displaystyle 9!\), the \(\displaystyle 9!\) will cancel out leaving:

\(\displaystyle \frac{12\times 11\times 10}{3\times 2\times 1}\)

\(\displaystyle \frac{12\times 11\times 10}{3\times 2\times 1} = \frac{1320}{6} = 220\)

Step 1: Evaluate \(\displaystyle 10!\)

\(\displaystyle 10!=10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=3628800\)

Step 2: Evaluate \(\displaystyle 4(9!)\)

\(\displaystyle 4(9!)=4(9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1)=1451520\)

Step 3: Evaluate \(\displaystyle 8!\)

\(\displaystyle 8!=8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=40320\)

Step 4: Evaluate \(\displaystyle 3(7!)\)

\(\displaystyle 3(7!)=3(7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1)=15120\)

Step 5: Add/Subtract values:

\(\displaystyle 3628800-1451520+40320-15120=2202480\)

Step 1: Do the division of the factorials

\(\displaystyle \bigg[\frac {9!}{7!}\bigg]=\frac {9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}\)

Step 2: Simplify by cancelling out any terms on both top and bottom:

\(\displaystyle \frac {9\cdot 8\cdot {\color{Blue} 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}{{\color{Blue} 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}=9\cdot8=72\)

Step 3: Multiply the result of the division by 3..

\(\displaystyle 72 \cdot 3=216\)

The value of that expression is \(\displaystyle 216\).

Step 1: Find \(\displaystyle 7!\)

\(\displaystyle 7!=7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=5040\)

Step 2: Find \(\displaystyle 6(6!)\)

\(\displaystyle 6(6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1)=6(720)=4320\)

Step 3: Subtract the values from step 1 and step 2:

\(\displaystyle 5040-4320=720=6!\)

The answer is \(\displaystyle 6!\)