The number of ways in which 10 persons can go | vedclass

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(B) Let the two particular persons be $P_1$ and $P_2$.Since $P_1$ and $P_2$ cannot be in the same boat,$P_1$ must be in one boat and $P_2$ must be in

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$2(^8C_4)$

Vedclass · Answer

(B) Let the two particular persons be $P_1$ and $P_2$.Since $P_1$ and $P_2$ cannot be in the same boat,$P_1$ must be in one boat and $P_2$ must be in the other.Each boat needs $5$ persons in total. Since $P_1$ is already in the first boat,we need to choose $4$ more persons from the remaining $8$ persons to join $P_1$.This can be done in $^8C_4$ ways.The remaining $4$ persons will automatically go into the second boat with $P_2$.Since the boats are distinct (or the arrangement of $P_1$ and $P_2$ can be swapped between the two boats),we multiply by $2$.Total number of ways $= 2 	imes ^8C_4$.

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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