Question

A lady desires to give a dinner party for 8 guests. In how many ways can the lady select guests for the dinner from her 12 friends, if two of the guests will not attend the party together?

Answer 1

The following three methods of approach are indicated.

(a) Number of ways of forming the party = ¹²C₈ - ¹⁰C₆

[since ¹⁰C₆ is the number of ways of making up the party with both the specified guests included]

= 495 - 210 = 285

(OR)

(b) Number of ways of forming the party

= Number of ways of forming without both of them + Number of ways of forming with one of them and without the other

=  ¹⁰C₈ + 2. ¹⁰C₇ = 45 + 240 = 285

(OR)

(c) Split the number of ways of forming the party

= Those with one of the two (say A) + those without A

=  =  ¹⁰C₇ +  ¹¹C₈  = 120 + 165 = 285