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Question Number 105625 by  M±th+et+s last updated on 30/Jul/20
prove that   ((m),(m) ) (((   n)),((n−k)) )+ (((    m)),((m−1)) ) (((       n)),((n−k+1)) )  + (((   m)),((m−2)) ) (((        n)),((n−k+2)) )+......+ (((    m)),((m−k)) ) ((n),(n) )  = (((m+n)),((     k)) )
$${prove}\:{that} \\ $$$$\begin{pmatrix}{{m}}\\{{m}}\end{pmatrix}\begin{pmatrix}{\:\:\:{n}}\\{{n}−{k}}\end{pmatrix}+\begin{pmatrix}{\:\:\:\:{m}}\\{{m}−\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\:\:\:\:\:\:\:{n}}\\{{n}−{k}+\mathrm{1}}\end{pmatrix} \\ $$$$+\begin{pmatrix}{\:\:\:{m}}\\{{m}−\mathrm{2}}\end{pmatrix}\begin{pmatrix}{\:\:\:\:\:\:\:\:{n}}\\{{n}−{k}+\mathrm{2}}\end{pmatrix}+……+\begin{pmatrix}{\:\:\:\:{m}}\\{{m}−{k}}\end{pmatrix}\begin{pmatrix}{{n}}\\{{n}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{{m}+{n}}\\{\:\:\:\:\:{k}}\end{pmatrix} \\ $$
Answered by prakash jain last updated on 30/Jul/20
(1+x)^(m+n) =(1+x)^m (1+x)^n   compare coefficient of x^k  on LHS  and RHS.  LHS=C_k ^(m+n)   RHS  (C_0 ^m x^m +C_1 ^m x^(m−1) +..+C_(m−k) ^m x^k +..+C_m ^m )×  (C_0 ^n x^n +C_1 ^n x^(n−1) +..+C_(n−k) ^n x^k +..+C_n ^n )×  coefficient of x^k   C_m ^m C_(n−k) ^n +..+C_n ^n C_(n−k) ^n
$$\left(\mathrm{1}+{x}\right)^{{m}+{n}} =\left(\mathrm{1}+{x}\right)^{{m}} \left(\mathrm{1}+{x}\right)^{{n}} \\ $$$$\mathrm{compare}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{k}} \:\mathrm{on}\:\mathrm{LHS} \\ $$$$\mathrm{and}\:\mathrm{RHS}. \\ $$$$\mathrm{LHS}=\mathrm{C}_{{k}} ^{{m}+{n}} \\ $$$$\mathrm{RHS} \\ $$$$\left({C}_{\mathrm{0}} ^{{m}} {x}^{{m}} +{C}_{\mathrm{1}} ^{{m}} {x}^{{m}−\mathrm{1}} +..+{C}_{{m}−{k}} ^{{m}} {x}^{{k}} +..+{C}_{{m}} ^{{m}} \right)× \\ $$$$\left({C}_{\mathrm{0}} ^{{n}} {x}^{{n}} +{C}_{\mathrm{1}} ^{{n}} {x}^{{n}−\mathrm{1}} +..+{C}_{{n}−{k}} ^{{n}} {x}^{{k}} +..+{C}_{{n}} ^{{n}} \right)× \\ $$$${coefficient}\:{of}\:{x}^{{k}} \\ $$$${C}_{{m}} ^{{m}} {C}_{{n}−{k}} ^{{n}} +..+{C}_{{n}} ^{{n}} {C}_{{n}−{k}} ^{{n}} \\ $$

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