3 items x y and z will have 6 different

Question

3 items x, y and z will have 6 different permutations however only one combination. The given formular is generally used to determine the number of combinations in a described situation.

ⁿC_r = (n!)/((r!)(n - r)!)

 Solution

i. ⁸C₇ = (8!)/((7!)(8 - 7)!)

= 8!/7! 1!

= (8 * 7!)/1 * 7!

= 8

ii. ⁶C₄ = (6!)/((4!)(6 - 4)!)

= 6!/4! 2!

= (6 * 5 * 4!)/(4! * 2 *1)

= 15

iii. 8C₃ = (8!)/((3!)(8 - 3)!)

= (8 *7 *6 *5!)/(3 * 2 * 1 * 5!)

= 56