0

$16$ guest have to be seated around $2$ circular tables,each accommodating 8members.$3$ particular guest desire to sit at one particular table and $4$ others at the other table.The number of ways of arranging these guest is:

options

a) $9C5$ b) $9!$$(7!)^2$/$4!$$5!$ c) $9C5$ . $4C4$ d) None of these

MyApproach:

After arranging $3$ guest and $4$ guest on each side respectively,we are left with $9$ people.Then,we select $5$ people from Ist Side and Now we are left with $4$ people.

Thus,remaining $4$ people can be arranged in $4!$ ways.

$9C5$ . $4C4$ . $4!$ . $5!$ Ans

Can Anyone give me the hint Why I am wrong?

Jack
  • 752

1 Answers1

0

You forgot to actually arrange the $3$ and $4$ "special" people in $2!3!$ ways

I get $\dbinom{9}{5}2!3!4!5!$ which would correspond to option $(d)$

[You can add $\dbinom{4}{4}$ to the expression if you like, but it is redundant ]

true blue anil
  • 41,295