How many ways can 4 people be chosen from a group of 9

tell whether the situation is permutation or combination.the solve
please i need help with this question
anyone help

C(n, r) = n! / (n - r)!(r!)

n = 9
r = 4

! means factorial

9! = 9*8*7*6*5*4*3*2*1
4! = 4*3*2*1
5! = 5*4*3*2*1

9!/4!(5!) = 126

Each person is an individual. Once you've picked a person they can't be picked again - that is, there can be no repetition. Further, the ORDER that the four people are picked in doesn't matter - this makes it a combination, rather than a permutation.

So, the formula for a combination without repetition, nCr (n people choose r, or in your case, 9 people choose 4):

n! / (n-r)! * r! where n = 9, r = 4

9! / 5! * 4! = (9*8*7*6) / 4*3*2*1 = 3024 / 24 = 126

There are 126 ways that four people can be chosen from a group of 9.

four people chosen from a group of 9
in case, order would not matter.
therefore it is a combination problem
C(n, r) = n! / (n - r)!(r!)

n = 9
r = 4
C(9, 4)
= 9! / (9 - 4)!(4!)
= 9! / (5! 4!)
= 126

Choosing involves Combinations.

Number of ways to select 4 people out of 9 is...

⁹C₄ = 9! / (4! x 5!) = (9 x 8 x 7 x 6) / (4 x 3 x 2) = 126

126 through factorials, I believe it's a combination.
Combination = Order is irrelevant
Permutation = Order is relevant (Padlock code)

its not specific .i think ..better way is forget number