combinations and permutations

Question 1

what are permutations

Answer 1

choosing subsets when order matters (e.g. combination to a safe)

Question 2

what are combinations

Answer 2

choosing subsets when order does not matter (e.g. lottery numbers)

Question 3

the two types of selection

Answer 3

With replacement (or repetition)

Without replacement

Question 4

permutations with replacement: Example Choose 2 objects from 4 objects - A, B, C, D

Answer 4

16 possibilities
Number of permutations is 16=4×4

Question 5

formula for Permutations (with replacement)

Answer 5

𝑛^𝑟
where 𝑛 = total number of objects and 𝑟 = number of chosen objects

Question 6

Permutations without replacement: Example Choose 2 objects from 4 objects - A, B, C, D

Answer 6

12 possibilities
For each choice the number reduces: 4×3=12

Question 7

formula for Permutations without replacement

Answer 7

𝑛𝑃𝑟 = 𝑛!/(𝑛-r)!

Question 8

whst happens to the Permutations without replacement formula when
𝑟=𝑛

Answer 8

𝑛𝑃𝑟 = 𝑛!

Question 9

Combinations (without replacement): Example Choose any 2 from 4 objects - A, B, C, D where the order doesn’t matter.

Answer 9

6 possibilities

Question 10

formula for Combinations (without replacement)

Answer 10

𝑛C𝑟 = 𝑛!/r!(𝑛-r)!

Question 11

Combinations (with replacement)
Example Choose any 2 from 4 objects where the order doesn’t matter - A, B, C, D

Answer 11

10 possibilities

Question 12

Combinations (with replacement) formula

Answer 12

(r + n-1)! / (n-1)!r!

Question 13

How do you calculate the probability of winning a lottery where you choose 6 numbers from 49?

Answer 13

The number of ways to choose 6 numbers from 49 (combination formula):
49!/(49−6)!6! = 13,983,816

The probability for any one set of numbers is:
1 / 13,983,816