JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 19)
If a, b and c are the greatest value of 19Cp, 20Cq
and 21Cr respectively, then :
$${a \over {11}} = {b \over {22}} = {c \over {21}}$$
$${a \over {10}} = {b \over {22}} = {c \over {21}}$$
$${a \over {10}} = {b \over {11}} = {c \over {42}}$$
$${a \over {11}} = {b \over {22}} = {c \over {42}}$$
Explanation
(
19Cp)max =
19C9 or
19C10 = a
( 20Cp)max = 20C10 = b
( 21Cr)max = 21C10 or 21C11 = c
1 = $${a \over {^{19}{C_{10}}}} = {b \over {^{20}{C_{10}}}} = {c \over {^{21}{C_{10}}}}$$
$$ \Rightarrow $$ $${a \over {11}} = {b \over {22}} = {c \over {42}}$$
( 20Cp)max = 20C10 = b
( 21Cr)max = 21C10 or 21C11 = c
1 = $${a \over {^{19}{C_{10}}}} = {b \over {^{20}{C_{10}}}} = {c \over {^{21}{C_{10}}}}$$
$$ \Rightarrow $$ $${a \over {11}} = {b \over {22}} = {c \over {42}}$$
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