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,one must be in the first boat and the other in

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$2(^8C_4)$

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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,one must be in the first boat and the other in the second boat.If $P_1$ is in the first boat,we need to select $4$ 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 to the second boat with $P_2$.Alternatively,if $P_2$ is in the first boat,we select $4$ persons from the remaining $8$ to join $P_2$,which can be done in $^8C_4$ ways.Thus,the total number of ways is $^8C_4 + ^8C_4 = 2 	imes ^8C_4$.

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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