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Using binomial theorem (without using the formula for `^n C_r` ) , prove that `^n C_4+^m C_2-^m C_1^n C_2=^m C_4-^(m+n)C_1^m C_3+^(m+n)C_2^m C_2-^(m+n)C_3^m C_1+^(m+n)C_4dot`

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`.^(m)C_(4)-.^(m+n)C_(1).^(m)C_(3)+.^(m+n)C_(2).^(m)C_(2)-.^(m+n)C_(3).^(m)C_(1)+.^(m+n)C_(4)`
`= .^(m+n)C_(0).^(m)C_(4)-.^(m+n)C_(1).^(m)C_(3)+.^(m+n)C_(2).^(m)C_(2)-.^(m+n)C_(3).^(m)C_(1)+.^(m+n)C_(4).^(m)C_(0)`
= Coefficient of `x^(4)` in `(1+x)^(m+n)(1-x)^(m)`
= Coefficient of `x^(4)` in `(1-x)^(m)(1+x)^(n)`
= Coefficient of `x^(4)` in `[1-.^(m)C_(1)x^(2)+.^(m)C_(2)x^(4)-"....."][1+.^(n)C_(1)x+.^(n)C_(2)x^(2)+"....."+.^(n)C_(n)x^(n)]`
`= .^(n)C_(4)-.^(m)C_(1) xx .^(n)C_(2)+.^(m)C_(2)`
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