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Question: If \(m{\text{ and }}p\) are positive \((m \geqslant p)\) and \(\Delta (m,p) = \left| {\begin{array}{...

If m and pm{\text{ and }}p are positive (m⩾p)(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 mCp=0 if m<p{}^m{C_p} = 0{\text{ if }}m < p, then
This equation has multiple correct options:
A. Δ(2,1)/Δ(1,0)=4\Delta (2,1)/\Delta (1,0) = 4
B. Δ(4,3)/Δ(3,2)=2\Delta (4,3)/\Delta (3,2) = 2
C. Δ(4,3)/Δ(2,1)=5\Delta (4,3)/\Delta (2,1) = 5
D. Δ(4,3)/Δ(1,0)=10\Delta (4,3)/\Delta (1,0) = 10

Explanation

Solution

If we are given nCr{}^n{C_r} then it means that nCr=n!(n−r)!r!{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}} and here !! means factorial for the number. Also we should know that nCr=nCn−r{}^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 mCp=0 if m<p{}^m{C_p} = 0{\text{ if }}m < p
Now for m>pm > p
mCp=m!(m−p)!p!{}^m{C_p} = \dfrac{{m!}}{{(m - p)!p!}}
So for option A
We need to find Δ(2,1),Δ(1,0)\Delta (2,1),\Delta (1,0)
So we get that m=2,p=1m = 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 nCn=1{}^n{C_n} = 1
And we know that for mCp if m<p then mCp=0 here 2C3=0 {}^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 4C2=4!2!2!=4(3)2(1)=6,2C3=0{}^4{C_2} = \dfrac{{4!}}{{2!2!}} = \dfrac{{4(3)}}{{2(1)}} = 6,{}^2{C_3} = 0
So we get that Δ(2,1)=2(12−6)−1(12−4)+0(18−12)\Delta (2,1) = 2(12 - 6) - 1(12 - 4) + 0(18 - 12)
Δ(2,1)=2(6)−8 =4  \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!=1,1C0=1!1!0!=1,2C0=2!(2−0)!0!=10! = 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 Δ(2,1)/Δ(1,0)=4/1=4\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 Δ(4,3),Δ(3,2)\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  if m<p then mCp=0 here 4C5=0 {\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 Δ(4,3)=4(30−15)−1(150−100)+0=60−50=10\Delta (4,3) = 4(30 - 15) - 1(150 - 100) + 0 = 60 - 50 = 10
Now forΔ(3,2)\Delta (3,2), m=3,p=2m = 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  if m<p then mCp=0 here 3C4=0 {\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 Δ(3,2)=3(20−10)−1(30−10)+0=10\Delta (3,2) = 3(20 - 10) - 1(30 - 10) + 0 = 10
So we get that Δ(4,3)=10\Delta (4,3) = 10 and Δ(3,2)=10\Delta (3,2) = 10
So Δ(4,3)/Δ(3,2)=10/10=1\Delta (4,3)/\Delta (3,2) = 10/10 = 1
Hence option B is also incorrect.
For option C
We know thatΔ(4,3)=10\Delta (4,3) = 10, Δ(2,1)=4\Delta (2,1) = 4
So Δ(4,3)/Δ(2,1)=10/4=2.5\Delta (4,3)/\Delta (2,1) = 10/4 = 2.5
Hence option C is not correct
For option D
We know thatΔ(4,3)=10\Delta (4,3) = 10,Δ(1,0)=1\Delta (1,0) = 1
Δ(4,3)/Δ(1,0)=10/1=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).........3.2.1n! = n(n - 1)(n - 2).........3.2.1 and we know that n>0n > 0
If n=0n = 0 then 0!=10! = 1
Similarly for the combination we know that nCr=n!(n−r)!r!{}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}
And for permutation we know that nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}
Combination is the number of ways of choosing rr things whereas permutation includes the selection as well as the arrangement of r out of nr{\text{ out of }}n things.