Question

A class has 30 students. In how many ways can committees be forced using the following numbers of students?
a. 3 students
b. 5 students

Answer 1

these is the combination problem. just use the formula :
nCr = n!/(n-r)!r!
in this case n = 30

part a, given r = 3
so, it is 30C3
= 30!/(30-3)3! = 30!/27!3! = 30 * 29 * 28 * 27!/(27! * 3 * 2 * 1) = 4060 ways.

now can you get part b ?

Answer 2

im writing the part a one second :))

Answer 3

another 'shortcut' way to work these out is for example for a@

number if combnations = 30 * 29 * 28
----------
3 * 2 * 1

Answer 4

b would be

30 * 29 *28*27*26
---------------
5 *4 * 3 * 2 * 1

Answer 5

im lost on part a how you got 4060

Answer 6

30 * 29 * 28 * 27!/(27! * 3 * 2 * 1)
cancel the 27!'s, get
30 * 29 * 28 / 3 * 2 * 1 = 24360/6 = 4060

part b. just put r = 5 into that formula. mom :)

Answer 7

that's basically what I did - I just cancelled the 27! in my head
in part b - you cancel the 25!

Answer 8

do you see the pattern in my calculation?
you can then cancel out the numbers further to make the calculation easier: