A certain lottery has 36 numbers. In how many different ways can 4 of the numbers be​ selected? (Assume that order of selection is not​ important.)

Question

A certain lottery has

36

numbers. In how many different ways can

44

of the numbers be​ selected? (Assume that order of selection is not​ important.)

Answer 1

this question is an example of a combination problem because the order of selection is not important! you can compute it by taking

C(36, 4) = 36! / 4!(36 - 4)! = 36! / 4!32! = 58,905

meaning that there are 58,905 different ways to select 4 numbers out of 36

Answer 2

This is a combination problem since the order does not matter. We use the notation ₃₆ C ₄ read as a combination of "36 choose 4"

The number of combinations of 4 numbers can be calculated using a function on calculator: TI 84 you would enter 36, hit MATH, then select PROB, then _nC_r, hit enter, then enter 4 and the number of combinations will be displayed.

by hand: you will use the formula n! / (n-r)! r!

n= number available

r= number chosen

a factorial is 1x2x3x...x n

so for this problem: 36! / (36-4)! 4! = (36!/32! ) x (1/4!)

36!/32! will be 33x34x35x36 = 1,413,720

4! = 24

so final answer is 124,740/24 = 58,905

Answer 3

To determine the number of ways to choose 4 numbers from a set of 36 when the order does not matter, we use the combination formula.

where:

1. n is the total number of items (36 numbers),
2. r is the number of items to choose (4 numbers).

After substituting in the values and crunching the numbers (I would use a calculator), there are 58,905 different ways to select 4 numbers from 36.

Hope this was helpful.