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Home  >>  EAMCET  >>  Mathematics
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$^{15}P_{8}=A+8.^{14}P_7 =>A=$

\[\begin {array} {1 1} (a)\;^{14}P_6 & \quad (b)\;^{14}P_8 \\ (c)\;^{15}P_7 & \quad (d)\;^{16}P_9 \end {array}\]
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$(b)\;^{14}P_8$
answered Nov 7, 2013 by pady_1
 

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