Permutations and, combinations: problem type 2.

(A) From a collection of 51 store customers, 3 are to be chosen to receive a special gift. How many
groups of 3 customers are possible?

(B) How, many different committees of size 4 can be formed from 14 people ?

Respuesta :

Step-by-step explanation:

Permutations:

nP r = n!/(n + r)!

Combinations:

nCr = n!/(n - r)!r!

c - combination

p - permutations

Since in both the problems they are looking for different groups, it is a combination problem. the order in which the people were selected in each group is not important, what is required are different groups.

A.

n = 51

r = 3

51C 3 = 51! / ((51 - 3) ×3!)

51C3 = 20 825 groups

OR:

[tex]c = \frac{51}{3} \times \frac{50}{2} \times 49[/tex]

[tex]c = 20825[/tex]

B.

[tex]c = \frac{14}{4} \times \frac{13}{3} \times \frac{12}{2} \times 11[/tex]

[tex]c = 1001[/tex]