Question

17．計(jì)算：C${\;}_{4}^{3}$+C${\;}_{5}^{3}$+…+C${\;}_{10}^{3}$=329．

Answer 1

分析 利用組合數(shù)公式${C}_{n}^{m+1}$+${C}_{n}^{m}$=${C}_{n+1}^{m+1}$進(jìn)行化簡(jiǎn)即可．

解答 解：C${\;}_{4}^{3}$+C${\;}_{5}^{3}$+…+C${\;}_{10}^{3}$=（${C}_{4}^{4}$+${C}_{4}^{3}$）+${C}_{5}^{3}$+…+${C}_{10}^{3}$-1
=${C}_{5}^{4}$+${C}_{5}^{3}$+…+${C}_{10}^{3}$-1
=${C}_{6}^{4}$+…+${C}_{10}^{3}$-1
=${C}_{10}^{4}$+${C}_{10}^{3}$-1
=${C}_{11}^{4}$-1
=329．
故答案為：329．

點(diǎn)評(píng) 本題考查了組合數(shù)公式的應(yīng)用問(wèn)題，也考查了公式靈活應(yīng)用的問(wèn)題，是基礎(chǔ)題目．