# Question 357c1

Apr 3, 2017

""_32 C_8 = 10,518,300 combinations

#### Explanation:

The form of a combination is ""_nC_r, where $n$ = the total number and $r =$the subset.

""_32 C_8 = (n!)/(r!(n-r)!) = (32!)/(8! (32-8)!) = (32!)/(8!*24!)

=(32*31*30*29*28*27*26*25*24!)/(8*7*6*5*4*3*2*1*24!)

The 24! cancels:

$= \frac{32 \cdot 31 \cdot 30 \cdot 29 \cdot 28 \cdot 27 \cdot 26 \cdot 25}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}$

Factor: $= \frac{8 \cdot 4 \cdot 31 \cdot 5 \cdot 6 \cdot 29 \cdot 7 \cdot 2 \cdot 2 \cdot 3 \cdot 9 \cdot 26 \cdot 25}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}$

Cancel numbers common in the numerator & denominator:
$= 31 \cdot 29 \cdot 2 \cdot 9 \cdot 26 \cdot 25$

$= 10 , 518 , 300$ combinations

If you have a TI-83, 84 calculator, you can type
$32$ MATH $\to P R B \text{ "3 " } 8$ ENTER

Apr 3, 2017

The result is 10 518 300 ways.

#### Explanation:

To calculate the number of ways that groups of eight can be chosen from a larger group of 32, we use the formula

""_nC_r = (n!)/((r!)(n-r)!)

So, in this case

""_32C_8 = (32!)/(8!*24!) = (32xx31xx30xx29xx28xx27xx26xx25)/(8xx7xx6xx5xx4xx3xx2xx1)

= 10 518 300

By the way, I believe the accepted notation is ""_nC_r#