Question

If ‘m’ and ‘p’ are positive $(m \geqslant p)$ and \Delta (m,p) = \left| {\begin{array}{*{20}{c}} {{}^m{C_p}}&{{}^m{C_{p + 1}}}&{{}^m{C_{p + 2}}} \\\ {{}^{m + 1}{C_p}}&{{}^{m + 1}{C_{p + 1}}}&{{}^{m + 1}{C_{p + 2}}} \\\ {{}^{m + 2}{C_p}}&{{}^{m + 2}{C_{p + 1}}}&{{}^{m + 2}{C_{p + 2}}} \end{array}} \right|
and ${}^m{C_p} = 0$ if $m < p$ , then
This question has multiple correct options.
A. $\Delta (2,1)/\Delta (1,0) = 4$
B. $\Delta (4,3)/\Delta (3,2) = 2$
C. $\Delta (4,3)/\Delta (2,1) = 5$
D. $\Delta (4,3)/\Delta (1,0) = 10$

Answer 1

Hint : For solving this particular question, we have to put $p = m - 1$ , then we have to simplify the given combinations as the number of combinations of $n$ different things taken $r$ at a time is given by ${}^n{C_r}$. Then, ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ .

Complete step by step solution:
It is given that ‘m’ and ‘p’ are positive $(m \geqslant p)$ and \Delta (m,p) = \left| {\begin{array}{*{20}{c}} {{}^m{C_p}}&{{}^m{C_{p + 1}}}&{{}^m{C_{p + 2}}} \\\ {{}^{m + 1}{C_p}}&{{}^{m + 1}{C_{p + 1}}}&{{}^{m + 1}{C_{p + 2}}} \\\ {{}^{m + 2}{C_p}}&{{}^{m + 2}{C_{p + 1}}}&{{}^{m + 2}{C_{p + 2}}} \end{array}} \right|
and ${}^m{C_p} = 0$ if $m < p$ ,
Now, consider $p = m - 1$ , we will get the following result ,
\Delta (m,m - 1) = \left| {\begin{array}{*{20}{c}} {{}^m{C_{m - 1}}}&{{}^m{C_m}}&{{}^m{C_{m + 1}}} \\\ {{}^{m + 1}{C_{m - 1}}}&{{}^{m + 1}{C_m}}&{{}^{m + 1}{C_{m + 1}}} \\\ {{}^{m + 2}{C_{m - 1}}}&{{}^{m + 2}{C_m}}&{{}^{m + 2}{C_{m + 1}}} \end{array}} \right|
Now, we know that the number of combinations of $n$ different things taken $r$ at a time is given by ${}^n{C_r}$ .
Then, ${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Therefore, we will get ,
\Delta (m,m - 1) = \left| {\begin{array}{*{20}{c}} m&1&0 \\\ {\dfrac{{(m + 1)m}}{2}}&{(m + 1)}&1 \\\ {\dfrac{{(m + 2)(m + 1)m}}{6}}&{\dfrac{{(m + 2)(m + 1)}}{2}}&{(m + 2)} \end{array}} \right|
= m(m + 2)\left| {\begin{array}{*{20}{c}} 1&1&0 \\\ {\dfrac{{(m + 1)}}{2}}&{(m + 1)}&1 \\\ {\dfrac{{(m + 1)}}{6}}&{\dfrac{{(m + 1)}}{2}}&1 \end{array}} \right|
$= \dfrac{{m(m + 1)(m + 2)}}{6}$
Now, we will check for the correct options,
Taking option ‘A’ $\Delta (2,1)/\Delta (1,0) = 4$
Now, according to the result ,
$\Delta (m,m - 1) = \dfrac{{m(m + 1)(m + 2)}}{6}$
$\Delta (2,1) = \dfrac{{2(2 + 1)(2 + 2)}}{6} = 4$ and
$\Delta (1,0) = \dfrac{{1(1 + 1)(1 + 2)}}{6} = 1$
Since $\Delta (2,1)/\Delta (1,0) = 4$
We can say that option ‘A’ is correct.

Now, taking option ‘B’ $\Delta (4,3)/\Delta (3,2) = 2$
Now, according to the result ,
$\Delta (m,m - 1) = \dfrac{{m(m + 1)(m + 2)}}{6}$
$\Delta (4,3) = \dfrac{{4(4 + 1)(4 + 2)}}{6} = 20$ and
$\Delta (3,2) = \dfrac{{3(3 + 1)(3 + 2)}}{6} = 10$
Since $\Delta (4,3)/\Delta (3,2) = 2$
We can say that option ‘B’ is also correct.

Now, taking option ‘C’ $\Delta (4,3)/\Delta (2,1) = 5$
Now, according to the result ,
$\Delta (m,m - 1) = \dfrac{{m(m + 1)(m + 2)}}{6}$
$\Delta (4,3) = \dfrac{{4(4 + 1)(4 + 2)}}{6} = 20$ and
$\Delta (2,1) = \dfrac{{2(2 + 1)(2 + 2)}}{6} = 4$
Since $\Delta (4,3)/\Delta (2,1) = 5$
We can say that option ‘C’ is also correct.

Now, taking option ‘D’ $\Delta (4,3)/\Delta (1,0) = 10$
Now, according to the result ,
$\Delta (m,m - 1) = \dfrac{{m(m + 1)(m + 2)}}{6}$
$\Delta (4,3) = \dfrac{{4(4 + 1)(4 + 2)}}{6} = 20$ and
$\Delta (1,0) = \dfrac{{1(1 + 1)(1 + 2)}}{6} = 1$
Since $\Delta (4,3)/\Delta (1,0) = 20$
We can say that option ‘D’ is incorrect.
Therefore, options ‘A’ , ‘B’, and ‘C’ are the correct options.
So, the correct answer is “OptionA,B and C”.

Note : We must know that determinant of the matrix of order $1 \times 1$ is given as $\Delta A = \left| A \right| = A$ , whereas the Determinant of matrices of order $2 \times 2$ is given as follow:
we have to consider two equations

ax + by = 0...(i)\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\; \\\ \;cx + dy = 0...(ii) \\\ \end{gathered} $$ Then , the result of $ \left| {\begin{array}{*{20}{c}} a&b; \\\ c&d; \end{array}} \right| $ is given as $ ad - bc. $