Question

In how many ways can $8$ persons be seated on two round tables of capacity $5$ and $3$.

Answer 1

Hint: To solve this question, we have to do it in $3$ steps. In step $1$, we have to select $5$ out of $8$ members to be seated on a big table, then in step $2$ arrange the $5$ members around the big table and then in step $3$ arrange the $3$ remaining members on the small table.

Complete step-by-step answer:
Step $1$: We have selected $5$ out of $8$ members to be seated on a big table.
$\therefore $No. of ways $ = {}^8{C_5}$
${
   = \dfrac{{8!}}{{5!3!}} \\
   = \dfrac{{8 \times 7 \times 6 \times 5!}}{{5!3 \times 2 \times 1}} \\
   = 56 \\
} $ [ according to law of combination, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$]
Step $2$: We arrange $5$ selected members around big round table whose capacity is 5 , so no. of ways in which it can be done
$\therefore $No. of ways $ = \left( {5 - 1} \right)!$
${
   = 4! \\
   = 4 \times 3 \times 2 \times 1 \\
   = 24 \\
} $
Step $3$: Now we arrange $3$ remaining members around small table whose capacity is $3$, so no. of ways it can be arranged is
$\therefore $No. of ways $ = \left( {3 - 1} \right)!$
${
   = 2! \\
   = 2 \\
} $
$\therefore $Total no. of ways $ = 56 \times 24 \times 2$$ = 2688$.
Hence, $8$ person can be seated around a $2$ different table in $2688$ ways.

Note: Permutation deals with the arrangement of items and combination deals with the selection of items. Order is important in permutations and order is not important in combinations.