The value of sum_{r=0}^4 {}^{(19-r)} C_3 + {} | vedclass

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(C) We use the identity ${}^{n} C_r + {}^{n} C_{r-1} = {}^{n+1} C_r$. Given expression: $\sum_{r=0}^4 {}^{19-r} C_3 + {}^{15} C_4$ $= {}^{19} C_3 + {}

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${}^{20} C_4$

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(C) We use the identity ${}^{n} C_r + {}^{n} C_{r-1} = {}^{n+1} C_r$. Given expression: $\sum_{r=0}^4 {}^{19-r} C_3 + {}^{15} C_4$ $= {}^{19} C_3 + {}^{18} C_3 + {}^{17} C_3 + {}^{16} C_3 + {}^{15} C_3 + {}^{15} C_4$ $= {}^{19} C_3 + {}^{18} C_3 + {}^{17} C_3 + {}^{16} C_3 + ({}^{15} C_3 + {}^{15} C_4)$ $= {}^{19} C_3 + {}^{18} C_3 + {}^{17} C_3 + ({}^{16} C_3 + {}^{16} C_4)$ $= {}^{19} C_3 + {}^{18} C_3 + ({}^{17} C_3 + {}^{17} C_4)$ $= {}^{19} C_3 + ({}^{18} C_3 + {}^{18} C_4)$ $= {}^{19} C_3 + {}^{19} C_4$ $= {}^{20} C_4$

The value of $\sum_{r=0}^4 {}^{(19-r)} C_3 + {}^{15} C_4$ is equal to

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