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Question

Question: In how many ways can \(8\) persons be seated on two round tables of capacity \(5\) and \(3\)....

In how many ways can 88 persons be seated on two round tables of capacity 55 and 33.

Explanation

Solution

To solve this question, we have to do it in 33 steps. In step 11, we have to select 55 out of 88 members to be seated on a big table, then in step 22 arrange the 55 members around the big table and then in step 33 arrange the 33 remaining members on the small table.

Complete step-by-step answer:
Step 11: We have selected 55 out of 88 members to be seated on a big table.
\therefore No. of ways =8C5 = {}^8{C_5}
=8!5!3! =8×7×6×5!5!3×2×1 =56 { = \dfrac{{8!}}{{5!3!}} \\\ = \dfrac{{8 \times 7 \times 6 \times 5!}}{{5!3 \times 2 \times 1}} \\\ = 56 \\\ } [ according to law of combination, nCr=n!r!(nr)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}]
Step 22: We arrange 55 selected members around big round table whose capacity is 5 , so no. of ways in which it can be done
\therefore No. of ways =(51)! = \left( {5 - 1} \right)!
=4! =4×3×2×1 =24 { = 4! \\\ = 4 \times 3 \times 2 \times 1 \\\ = 24 \\\ }
Step 33: Now we arrange 33 remaining members around small table whose capacity is 33, so no. of ways it can be arranged is
\therefore No. of ways =(31)! = \left( {3 - 1} \right)!
=2! =2 { = 2! \\\ = 2 \\\ }
\therefore Total no. of ways =56×24×2 = 56 \times 24 \times 2 =2688 = 2688.
Hence, 88 person can be seated around a 22 different table in 26882688 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.