Question

Evaluate ${}^{20}C_5+\sum _{r=2}^5{}^{25-r}{C}_4$.

Answer 1

Given:
${}^{20}{C}_5+\sum _{r=2}^{5}25{-}^r{C}_4$

$$\begin{gathered} {}^{20}{C}_5+\sum _{r=2}^{5}25{-}^r{C}_4 \\ ={}^{20}{C}_5{+}^{23}{C}_4+{}^{22}{C}_4+{}^{21}{C}_4+{}^{20}{C}_4 \\ =\left({}^{20}{C}_{4_{}}+{}^{20}{C}_5\right){+}^{21}{C}_4{+}^{22}{C}_4{+}^{23}{C}_4 \end{gathered}$$

${=}^{21}{C}_5{+}^{21}{C}_4+{}^{{}^{22}}{C}_4{+}^{23}{C}_4$ [∵ ${{}^{}}^n{C}_{r-1}{+}^n{C}_r{=}^{n+1}{C}_r$]
$=\left({}^{21}{C}_{4_{}}+{}^{21}{C}_{5_{}}\right){+}^{22}{C}_4{+}^{23}{C}_4$
${=}^{22}{C}_5{+}^{22}{C}_4{+}^{23}{C}_4$ [∵${{}^{}}^n{C}_{r-1}{+}^n{C}_r{=}^{n+1}{C}_r$]
${=}^{23}{C}_5{+}^{22}{C}_4{+}^{23}{C}_4$
$$\begin{gathered} {=}^{23}{C}_5{+}^{23}{C}_4 \\ {=}^{24}{C}_5 \end{gathered}$$ [∵ ${{}^{}}^n{C}_{r-1}{+}^n{C}_r{=}^{n+1}{C}_r$]

$=\frac{25!}{19!5!}=\frac{24\times 23\times 22\times 21\times 20}{5\times 4\times 3\times 2\times 1}=42504$