A person has 8 friends, of whom 5 will be invited to a party:a) How many choices are there if 2 of the friends are feudingand will not attend together?b) how many choices if 2 of the friends will only attendtogether?

Respuesta :

Answer:

a) 36

b) 26

Step-by-step explanation:

Data provided in the question:

Total friends = 8

Number of friends invited to the party = 5

Now,

a)  if 2 of the friends are feuding will not attend together

= ( Total choices of selecting 5 friends out of 8 ) - (Choices in which both feuding will attend)

= ⁸C₅ - ⁶C₃

[ here, when both will come we will be left with inviting 3 friends out of 6 as 2 are already fixed. thus  ⁶C₃ ]

= [tex]\frac{8!}{5!(8-5)!}-\frac{6!}{3!(6-3)!}[/tex]

= 36

b)  if 2 of the friends will only attend together

= Both attend the party + None of them attend the party

= ⁶C₃ + ⁶C₅

= [tex]\frac{6!}{3!(6-3)!}+\frac{6!}{5!(6-5)!}[/tex]

= 26

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