Permutation and combination regarding two people together You have 11 friends and can invite 5 to dinner 1) in how many ways if two of the friends married and will not attend separately? 2) in how many ways if two of them are not on speaking terms and will not attend together?

(1) as (1a) the two married plus 3 out of the remaining 9 plus (1b) 5 out of the 9 excluding the married.

(2) as (2a) one of the two plus 4 out of the remaining 9 (excluding 1 and her enemy) plus 2(b) the same again for the other of the two.

(1a) = 1. $^9C_3$ = $9!/3!.6!$ = 84

(1b) = $^9C_5$ = $9!/5!. 4!$ = 18

(1) = 102.

(2) = $2. ^9C_4$ = $2. 9!/5!.4!$ = 36.