How many different 5-person subcommittees can be formed from a club having 12 members

Given that there are 12 persons, the first choice may be in 12 different ways, the second choice may be in 11 different ways, ther third in 10 different ways, the fourth in 9 different ways and the fith in 8 different ways, for a total of:

12x11x10x9x8 different combinations.

Now you have to take in account that 5x4x3x2 are repetitions. So you have to divide the previos counting by 5x4x3x2.

(12x11x10x9x8)/(5x4x3x2) = 792 different subcommittees.

Also, you can use the formula for combinations: C(m,n) = m! / (n! (m-n)!)

C (12, 5) = 12! / (5!) (12-5)! = [12x11x10x9x8x7!] / [5! 7!] = [12x11x10x9x8]/[5x4x3x2] = 792