At the end of a cricket series,when five play | vedclass

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(C) $1$. Let the five players be $N, M, K, O, L$.$2$. For runs (ascending order): $N$ is $1^{st}$ and $O$ is $4^{th}$. Since $L$ has the $1^{st}$ rank

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(C) $1$. Let the five players be $N, M, K, O, L$.$2$. For runs (ascending order): $N$ is $1^{st}$ and $O$ is $4^{th}$. Since $L$ has the $1^{st}$ rank in one ranking and $5^{th}$ in another,and $N$ is $1^{st}$ in runs,$L$ must be $5^{th}$ in runs. Given $K > M$,the order is $N $3$. For wickets (descending order): The problem states $K$ replaces $O$ and $O$ replaces $L$ from the previous order,and $M$ remains unchanged. The original positions were $1:N, 2:M, 3:K, 4:O, 5:L$. $4$. Applying the replacements: $K$ takes $O$'s place $(4^{th})$,$O$ takes $L$'s place $(5^{th})$,and $M$ stays at $2^{nd}$. This leaves $N$ at $1^{st}$ and $L$ at $3^{rd}$.$5$. The descending order of wickets is $N > M > L > K > O$.$6$. Therefore,$O$ has taken the lowest number of wickets.

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