Question

I have 12 friends I would like to invite to dinner but can only accommodate ten. How many ways can I invite ten if: a. There are no restrictions. Simplify your answer to one final number. For b and c Do not simplify your answers. Leave in combinatorics form. b. Two of my friends are married so if I invite one, I have to invite the other. c. Two of my friends do not like each other so if invite one, I cannot invite the other.

Answer 1

there are 66 ways to invite 10 if there are no restrictions

Explanation:

a. to get it we use the combination rule

C(n,r)=?

(n,r)=(nr)=n!/(r!(n−r)!)=

C(n,r)=C(12,10)

=12!/(10!(12−10)!) = 12!/(10!2!)

= 12x11x10x9x8x7x6x5x4x3x2x1/(10x9x8x7x6x5x4x3x2x1(2x1)

= 79833600/(604800x2)

= 79833600/1209600 = 66