Question

The value of $\Delta =\begin{vmatrix} ^{10}C_4 & ^{10}C_5 & ^{11}C_m \\[0.3em] ^{11}C_6 & ^{11}C_7 &^{12}C_{m+2} \\[0.3em] ^{12}C_8 & ^{12}C_9 & ^{13}C_{m+4} \end{vmatrix}$ is equal to zero, where m is

• A. 6
• B. 4
• C. 5
• D. none of these.

Answer 1

Answer: (C)

$\Delta =\begin{vmatrix} ^{10}C_4 & ^{10}C_5 & ^{11}C_m \\[0.3em] ^{11}C_6 & ^{11}C_7 &^{12}C_{m+2} \\[0.3em] ^{12}C_8 & ^{12}C_9 & ^{13}C_{m+4} \end{vmatrix} = 0 $
Operate $C_1 + C_2$
$\Delta =\begin{vmatrix} ^{10}C_4 & ^{10}C_4+ ^{10}C_5 & ^{11}C_m \\[0.3em] ^{11}C_6 & ^{11}C_6+^{11}C_7 &^{12}C_{m+2} \\[0.3em] ^{12}C_8 & ^{12}C_8 +^{12}C_9 & ^{13}C_{m+4} \end{vmatrix} = 0 $
$=\begin{vmatrix} ^{10}C_4 & ^{12}C_4& ^{11}C_m \\[0.3em] ^{11}C_6 & ^{12}C_7 &^{12}C_{m+2} \\[0.3em] ^{12}C_8 &^{13}C_9& ^{13}C_{m+4} \end{vmatrix} = 0 $
Clearly, $m = 5$ satisfies the above result, [$\because \, C_1, C_3$ will be identical]