To solve the expression [tex]\(\frac{{ }_8 C_3}{{ }_7 C_6} - \frac{81!}{79!}\)[/tex], we need to look at each part individually and then combine the results.
### Step 1: Evaluate [tex]\({}_8 C_3\)[/tex]
The binomial coefficient [tex]\({}_8 C_3\)[/tex] (read as "8 choose 3") represents the number of ways to choose 3 items from a set of 8 without regard to order. The formula for combinations is:
[tex]\[
{}_n C_r = \frac{n!}{r!(n - r)!}
\][/tex]
Substituting [tex]\(n = 8\)[/tex] and [tex]\(r = 3\)[/tex]:
[tex]\[
{}_8 C_3 = \frac{8!}{3!(8 - 3)!} = \frac{8!}{3! \cdot 5!}
\][/tex]
From the problem, we know:
[tex]\[
{}_8 C_3 = 56
\][/tex]
### Step 2: Evaluate [tex]\({}_7 C_6\)[/tex]
The binomial coefficient [tex]\({}_7 C_6\)[/tex] (read as "7 choose 6") represents the number of ways to choose 6 items from a set of 7. For combinations:
[tex]\[
{}_7 C_6 = \frac{7!}{6!(7 - 6)!} = \frac{7!}{6! \cdot 1!}
\][/tex]
From the problem, we know:
[tex]\[
{}_7 C_6 = 7
\][/tex]
### Step 3: Compute the fraction [tex]\(\frac{{}_8 C_3}{{}_7 C_6}\)[/tex]
[tex]\[
\frac{{}_8 C_3}{{}_7 C_6} = \frac{56}{7} = 8.0
\][/tex]
### Step 4: Evaluate [tex]\(\frac{81!}{79!}\)[/tex]
Using the property of factorials, we recognize that:
[tex]\[
\frac{81!}{79!} = \frac{81 \cdot 80 \cdot 79!}{79!} = 81 \cdot 80
\][/tex]
From the problem, we know:
[tex]\[
\frac{81!}{79!} = 6480
\][/tex]
### Step 5: Combine the results
Now, we put together the parts:
[tex]\[
\frac{{ }_8 C_3}{{ }_7 C_6} - \frac{81!}{79!} = 8.0 - 6480
\][/tex]
Simplifying the expression:
[tex]\[
8.0 - 6480 = -6472
\][/tex]
Thus, the solution to the expression is:
[tex]\(\boxed{-6472}\)[/tex]
