Question #357c1

2 Answers
Apr 3, 2017

Answer:

#""_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

Apr 3, 2017

Answer:

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#