Question

Question #357c1

Answer 1

`""_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:

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

Factor: `=(8*4*31*5*6*29*7*2*2*3*9*26*25)/(8*7*6*5*4*3*2)`

Cancel numbers common in the numerator & denominator:
`=31*29*2*9*26*25`

`= 10,518,300` combinations

If you have a TI-83, 84 calculator, you can type
`32` MATH `-> PRB " "3 " "8` ENTER

Answer 2

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`