A shop sells 20 different flavors of ice cream, in how many ways can a customer choose4 ice cream cones (one dip of ice cream per cone, please) if they:(a) are all of different flavors.(b) are not necessarily of different flavors.(c) contain only 2 or 3 flavors.(d) contain 3 different flavors.

a.4845ways.

b. 14535ways.

c. 3990ways

d. 1140ways

Step-by-step explanation:

Given data:

No of flavors available to customers = 20.

Solution:

This is permutation and combinations problem,

(a) how many ways can the customers choose 4 different ice creams if they are all of different flavors.

20C4

= n!/(n-k)!)k!

= 20!/(20-4)!)4!

= 20!/(16)!)4!

= 4845ways.

b) are not necessarily of different flavors

Let’s say any two same flavors can be chosen.

20C4 * 3!/2!

= 4845 * 3

= 14535ways.

c) contain only 2 or 3 flavors.

= 20C3 * 3!/2!

= 1140 * 3

= 3420

20C2 * 3

= 190 * 3

= 570.

No of 2 or 3 different flavors

= 3420 + 570

= 3990ways.

d) contain 3 different flavors.

20C3

= n!/(n-k)!)k!

= 20!/(20-3)!)3!

= 1140ways.

the answer is d. (1.2, -4.7) on edgn

hope this !