A five-member review team is to be chosen from 6 senior staff and 5 junior staff. The team must contain more seniors than juniors. In how many ways can it be chosen?

With five places, 'more seniors than juniors' means 3, 4 or 5 seniors. Three seniors and two juniors: $C(6,3)\times C(5,2)=20\times 10=200$. Four seniors and one junior: $C(6,4)\times C(5,1)=15\times 5=75$. Five seniors and no junior: $C(6,5)=6$. Total = 200 + 75 + 6 = 281. Trap 275 drops the all-senior team, which is easy to overlook because it needs no junior at all. Trap 200 keeps only the 3–2 split. Trap $462=C(11,5)$ ignores the condition and counts every possible team.