A business man hosts a dinner to 21 guests. He is having 2 round tables which can accommodate 15 and 6 persons each. In how many ways can he arrange the guests ?

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Solution

In one round table the business man can accommodate the guests in $^{21} C_{1} 5$ ways. Keeping one guest as fixed in the first round table, the other 14 guests can be arrange in 14! ways. Keeping one guest as fixed in the second round table, the other 5 guests can be arrange in 5! ways.

Therefore the total number of ways in which the guests can be arrange is

$^{21} C_{1} 5 \times^{6} C_{6} \times 14! \times 5!$ ways.

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