A businessman 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

Out of $21,$ we can choose $15$ for one table in $^{21} C_{15}$ ways
And for each selection we are left with $6$ guests for the second table having $6$ seats.
This can be chosen in $^{6} C_{6}$ ways.
Now $15$ for round table $A$ can be arranged in $(14)!$ ways and $6$ for round table B can be arranged in $5!$ ways.
Hence the total number is $^{21} C_{15} \times^{6} C_{6} \times 14! \times 5!$ .

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