TG EAMCET 3 May 2025 Shift 1 Paper

∵ Total number of ways in which train has stopped at exactly 5 stations =‌15C5
And total number of ways in which train has not stopped in consecutive stations
x1+x2+x3+x4+x5+L=15−4=11
xi= number of station left before i‌th ‌ station
x1≥0x2,x3,x4,x5≥1, and L= station
left after 4 stations
x1+(x2−1)+(x3−1)+(x4−1)+(x5−1)+L
=11−4=7
∴‌ Total number of solution ‌‌=‌7+5−1C6−1
‌=‌11C5
∴ Required ways =‌15C5−‌11C5