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$^{15}P_{8}=A+8.^{14}P_7 =>A=$
\[\begin {array} {1 1} (a)\;^{14}P_6 & \quad (b)\;^{14}P_8 \\ (c)\;^{15}P_7 & \quad (d)\;^{16}P_9 \end {array}\]
jeemain
eamcet
math
2011
q24
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Sep 20, 2013
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$(b)\;^{14}P_8$
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Nov 7, 2013
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