There are five seniors in a class. for each situation, write how the binomial formula is used to calculate the probability.

a) in how many ways can i choose one senior to represent the group?
b) in how many ways can i choose two seniors to represent the group?
c) in how many ways can i choose three seniors to represent the group?
d) in how many ways can i choose four seniors to represent the group?
e) in how many ways can i choose five seniors to represent the group?

A) The answer is 5.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose one senior: r = 1

nCr = n! / (r! (n - r)!)
5C1 = 5! / (1! (5 - 1)!)
       = (5 * 4 * 3 * 2 * 1) / (1 * 4!)
       = 120 / (4 * 3 * 2 * 1)
       = 120 / 24
       = 5

b) The answer is 10.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose two seniors: r = 2

nCr = n! / (r! (n - r)!)
5C2 = 5! / (2! (5 - 2)!)
       = (5 * 4 * 3 * 2 * 1) / ((2 * 1) * 3!)
       = 120 / (2 * (3 * 2 * 1))
       = 120 / (2 * 6)
       = 120 / 12
       = 10

c) The answer is 10.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose three seniors: r = 3

nCr = n! / (r! (n - r)!)
5C3 = 5! / (3! (5 - 3)!)
       = (5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * 2!)
       = 120 / (6 * (2 * 1))
       = 120 / (6 * 2)
       = 120 / 12
       = 10

d) The answer is 5.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose four seniors: r = 4

nCr = n! / (r! (n - r)!)
5C4 = 5! / (4! (5 - 4)!)
       = (5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * 1!)
       = 120 / (24 * 1)
       = 120 / 24
       = 5

e) The answer is 1.

nCr = n! / (r! (n - r)!)
n - number of things to be chosen from
r - number of chosen things

There are five seniors in a class: n = 5
I choose five seniors: r = 5

nCr = n! / (r! (n - r)!)
5C5 = 5! / (5! (5 - 5)!)
       = (5 * 4 * 3 * 2 * 1) / ((5 * 4 * 3 * 2 * 1) * 1!)
       = 120 / (120 * 1)
       = 120 / 120
       = 1

He is right i did this thing before