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The number of ways in which 10 persons can go in two boats, so that there may be 5 on each boat, supposing that two particular persons will not go in the same boat is

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$\frac{1}{2} ({}^{10}C_{5})$

$\frac{1}{2} ({}^{8}C_{5})$

$2 \times {}^{8}C_{4}$

${}^{8}C_{4}$

answer is C.

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Detailed Solution

Two particular persons can be in different boats in 2 ways

Then the remaining 8 persons can be divided into two groups containing 4 each is $^{8} C_{4}$

ways

Therefore required number= $^{8} C_{4} \times 2$

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