Question

If $m{\text{ and }}p$ are positive $(m \geqslant p)$ and $\Delta (m,p) = \left| {\begin{array}{*{20}{l}}
  {{}^m{C_p}}&{{}^m{C_{p + 1}}}&{{}^m{C_{p + 2}}} \\
  {{}^{m + 1}{C_p}}&{{}^{m + 1}{C_{p + 1}}}&{{}^{m + 1}{C_{p + 2}}} \\
  {{}^{m + 2}{C_p}}&{{}^{m + 2}{C_{p + 1}}}&{{}^{m + 2}{C_{p + 2}}}
\end{array}} \right|$ and ${}^m{C_p} = 0{\text{ if }}m < p$, then
This equation 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:
If we are given ${}^n{C_r}$ then it means that ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ and here $!$ means factorial for the number. Also we should know that ${}^n{C_r} = {}^n{C_{n - r}}$

Complete step by step solution:
Here we are given that
$\Delta (m,p) = \left| {\begin{array}{*{20}{l}}
  {{}^m{C_p}}&{{}^m{C_{p + 1}}}&{{}^m{C_{p + 2}}} \\
  {{}^{m + 1}{C_p}}&{{}^{m + 1}{C_{p + 1}}}&{{}^{m + 1}{C_{p + 2}}} \\
  {{}^{m + 2}{C_p}}&{{}^{m + 2}{C_{p + 1}}}&{{}^{m + 2}{C_{p + 2}}}
\end{array}} \right|$
Now it is also given that ${}^m{C_p} = 0{\text{ if }}m < p$
Now for $m > p$
${}^m{C_p} = \dfrac{{m!}}{{(m - p)!p!}}$
So for option A
We need to find $\Delta (2,1),\Delta (1,0)$
So we get that $m = 2,p = 1$
$\Delta (2,1) = \left| {\begin{array}{*{20}{l}}
  {{}^2{C_1}}&{{}^2{C_{1 + 1}}}&{{}^2{C_{1 + 2}}} \\
  {{}^{2 + 1}{C_1}}&{{}^{2 + 1}{C_{1 + 1}}}&{{}^{2 + 1}{C_{1 + 2}}} \\
  {{}^{2 + 2}{C_1}}&{{}^{2 + 2}{C_{1 + 1}}}&{{}^{2 + 2}{C_{1 + 2}}}
\end{array}} \right|$
And we also know that ${}^n{C_n} = 1$
And we know that for ${}^m{C_p}{\text{ if }}m < p{\text{ then }}{}^m{C_p} = 0{\text{ here }}{}^2{C_3} = 0{\text{ }}$
$\Delta (2,1) = \left| {\begin{array}{*{20}{l}}
  {{}^2{C_1}}&{{}^2{C_2}}&{{}^2{C_3}} \\
  {{}^3{C_1}}&{{}^3{C_2}}&{{}^3{C_3}} \\
  {{}^4{C_1}}&{{}^4{C_2}}&{{}^4{C_3}}
\end{array}} \right| = \left| {\begin{array}{*{20}{l}}
  2&1&0 \\
  3&3&1 \\
  4&6&4
\end{array}} \right|$
We know that ${}^4{C_2} = \dfrac{{4!}}{{2!2!}} = \dfrac{{4(3)}}{{2(1)}} = 6,{}^2{C_3} = 0$
So we get that $\Delta (2,1) = 2(12 - 6) - 1(12 - 4) + 0(18 - 12)$
                           $
  \Delta (2,1) = 2(6) - 8 \\
   = 4 \\
 $
$\Delta (1,0) = \left| {\begin{array}{*{20}{l}}
  {{}^1{C_0}}&{{}^1{C_{0 + 1}}}&{{}^1{C_{0 + 2}}} \\
  {{}^{1 + 1}{C_0}}&{{}^{1 + 1}{C_{0 + 1}}}&{{}^{1 + 1}{C_{0 + 2}}} \\
  {{}^{1 + 1}{C_0}}&{{}^{1 + 2}{C_{0 + 1}}}&{{}^{1 + 2}{C_{0 + 2}}}
\end{array}} \right|$
             $ = \left| {\begin{array}{*{20}{l}}
  {{}^1{C_0}}&{{}^1{C_1}}&{{}^1{C_2}} \\
  {{}^2{C_0}}&{{}^2{C_1}}&{{}^2{C_2}} \\
  {{}^2{C_0}}&{{}^2{C_1}}&{{}^3{C_2}}
\end{array}} \right|$
So we get that $0! = 1,{}^1{C_0} = \dfrac{{1!}}{{1!0!}} = 1,{}^2{C_0} = \dfrac{{2!}}{{(2 - 0)!0!}} = 1$
So we get that $\Delta (1,0) = \left| {\begin{array}{*{20}{l}}
  1&1&0 \\
  1&2&1 \\
  1&3&3
\end{array}} \right|$
So we get that
So $\Delta (2,1)/\Delta (1,0) = 4/1 = 4$
Hence option A is right and similarly we need to check the other options also
For option B we need to find $\Delta (4,3),\Delta (3,2)$
$\Delta (4,3) = \left| {\begin{array}{*{20}{l}}
  {{}^4{C_3}}&{{}^4{C_4}}&{{}^4{C_5}} \\
  {{}^5{C_3}}&{{}^5{C_4}}&{{}^5{C_5}} \\
  {{}^6{C_3}}&{{}^6{C_4}}&{{}^6{C_5}}
\end{array}} \right|$
Now we know that ${\text{ if }}m < p{\text{ then }}{}^m{C_p} = 0{\text{ here }}{}^4{C_5} = 0{\text{ }}$
$\Delta (4,3) = \left| {\begin{array}{*{20}{l}}
  4&1&0 \\
  {\dfrac{{5!}}{{2!3!}}}&5&1 \\
  {\dfrac{{6!}}{{3!3!}}}&{\dfrac{{6!}}{{4!2!}}}&6
\end{array}} \right| = \left| {\begin{array}{*{20}{l}}
  4&1&0 \\
  {10}&5&1 \\
  {20}&{15}&6
\end{array}} \right|$
So we get that $\Delta (4,3) = 4(30 - 15) - 1(150 - 100) + 0 = 60 - 50 = 10$
Now for$\Delta (3,2)$, $m = 3,p = 2$
$\Delta (3,2) = \left| {\begin{array}{*{20}{l}}
  {{}^3{C_2}}&{{}^3{C_3}}&{{}^3{C_4}} \\
  {{}^4{C_2}}&{{}^4{C_3}}&{{}^4{C_4}} \\
  {{}^5{C_2}}&{{}^5{C_3}}&{{}^5{C_4}}
\end{array}} \right|$
Now we know that ${\text{ if }}m < p{\text{ then }}{}^m{C_p} = 0{\text{ here }}{}^3{C_4} = 0{\text{ }}$
$\Delta (3,2) = \left| {\begin{array}{*{20}{l}}
  3&1&0 \\
  {\dfrac{{4!}}{{2!2!}}}&4&1 \\
  {\dfrac{{5!}}{{2!3!}}}&{\dfrac{{5!}}{{3!2!}}}&5
\end{array}} \right| = \left| {\begin{array}{*{20}{l}}
  3&1&0 \\
  6&4&1 \\
  {10}&{10}&5
\end{array}} \right|$
So we get that $\Delta (3,2) = 3(20 - 10) - 1(30 - 10) + 0 = 10$
So we get that $\Delta (4,3) = 10$ and $\Delta (3,2) = 10$
So $\Delta (4,3)/\Delta (3,2) = 10/10 = 1$
Hence option B is also incorrect.
For option C
We know that$\Delta (4,3) = 10$, $\Delta (2,1) = 4$
So $\Delta (4,3)/\Delta (2,1) = 10/4 = 2.5$
Hence option C is not correct
For option D
We know that$\Delta (4,3) = 10$,$\Delta (1,0) = 1$
$\Delta (4,3)/\Delta (1,0) = 10/1 = 1$
Hence option D is also correct.

Hence we get that option A and D are correct.

Note:
We know that $n! = n(n - 1)(n - 2).........3.2.1$ and we know that $n > 0$
If $n = 0$ then $0! = 1$
Similarly for the combination we know that ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
And for permutation we know that ${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$
Combination is the number of ways of choosing $r$ things whereas permutation includes the selection as well as the arrangement of $r{\text{ out of }}n$ things.