HCU MCA 2013 Paper

(c) Case I ${\mathrm{n}}_1,{\mathrm{n}}_2,$=$\{\begin{array}{c}{\mathrm{n}}_3\\ {\mathrm{n}}_4\\ {\mathrm{n}}_5\\ {\mathrm{n}}_6\\ {\mathrm{n}}_7\\ {\mathrm{n}}_8\\ {\mathrm{n}}_9\\ {\mathrm{n}}_{10}\end{array}$ = 8 ways
Case II ${\mathrm{n}}_1,{\mathrm{n}}_3,$=$\{\begin{array}{c}{\mathrm{n}}_4\\ {\mathrm{n}}_5\\ {\mathrm{n}}_6\\ {\mathrm{n}}_7\\ {\mathrm{n}}_8\\ {\mathrm{n}}_9\\ {\mathrm{n}}_{10}\end{array}$ = 7 ways
Similarly, in this way we can find the total number of plant = [8 + 7 + 6 + 5 + 4+ 3 + 2 + 1]+ [7+6 + 5 + 4 + 3 + 2 + 1]+ [6 + 5+ 4 + 3 + 2 + 1] + [5 + 4 + 3 + 2+1]+ [4 + 3 + 2 + 1] + [3 + 2 +1] + [2 + 1] + [1]
= 36+28+ 21+ 15 + 10 + 6 + 3 + 1
=120