In the arrangement P, 2, 6, M, 9, K, S, 4, V, | vedclass

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(D) The original arrangement is: $P, 2, 6, M, 9, K, S, 4, V, 7$ (Total $10$ elements).The rule for interchange is: $1^{st} \leftrightarrow 6^{th}$,$2^

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(D) The original arrangement is: $P, 2, 6, M, 9, K, S, 4, V, 7$ (Total $10$ elements).The rule for interchange is: $1^{st} \leftrightarrow 6^{th}$,$2^{nd} \leftrightarrow 7^{th}$,$3^{rd} \leftrightarrow 8^{th}$,$4^{th} \leftrightarrow 9^{th}$,$5^{th} \leftrightarrow 10^{th}$.Applying these interchanges:$1^{st}$ $(P)$ and $6^{th}$ $(K)$ swap $\rightarrow K, 2, 6, M, 9, P, S, 4, V, 7$$2^{nd}$ $(2)$ and $7^{th}$ $(S)$ swap $\rightarrow K, S, 6, M, 9, P, 2, 4, V, 7$$3^{rd}$ $(6)$ and $8^{th}$ $(4)$ swap $\rightarrow K, S, 4, M, 9, P, 2, 6, V, 7$$4^{th}$ $(M)$ and $9^{th}$ $(V)$ swap $\rightarrow K, S, 4, V, 9, P, 2, 6, M, 7$$5^{th}$ $(9)$ and $10^{th}$ $(7)$ swap $\rightarrow K, S, 4, V, 7, P, 2, 6, M, 9$The new arrangement is: $K, S, 4, V, 7, P, 2, 6, M, 9$.The $4^{th}$ element from the left end is $V$.

In the arrangement $P, 2, 6, M, 9, K, S, 4, V, 7$,the positions of the $1^{st}$ and the $6^{th}$ elements are interchanged,those of the $2^{nd}$ and the $7^{th}$ are interchanged,and so on. Then,which of the following will be the $4^{th}$ element from the left end in the new arrangement?

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