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
A$\frac{1}{2} (^{10}C_5)$
B$\frac{1}{2} (^{8}C_{5})$
C$2 \times {}^{8}C_{4}$
D${}^{8}C_{4}$
Answer & Solution
Correct answer: D. ${}^{8}C_{4}$
Let the two particular persons be $A$ and $B$. Since they cannot be in the same boat and each boat has $5$ persons, one boat must contain $A$ and the other must contain $B$.
Now choose $4$ of the remaining $8$ persons to go with $A$. Then the other $4$ automatically go with $B$.
So the number of ways is
$$\binom{8}{4}$$
There is no further division by $2$ because once the group with $A$ is chosen, the group with $B$ is fixed, and $A$ and $B$ distinguish the two boats for counting.
On checking the options, this matches option $D$.
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