IanStewart wrote:
50 remaining ways to pick three people so that none are adjacent
Sir Stewart, your method is great. But can you tell me what's wrong in my reasoning?
Step 1: Pick any seat as the 1st winner-seat and name this seat as #1. Then two adjacent seats—#10 and #2—would be unavailable to be chosen.
Step 2: Pick #3 as the 2nd winner-seat. Then #4 would be unavailable. Now,
5 available seats are left, and they're #5, #6, #7, #8 and #9.
Step 3: Pick #4 as the 2nd winner-seat. Then #3 and #5 would be unavailable.
4 available seats are left, and they're #6, #7, #8 and #9.
Step 4: Pick #5 as the 2nd winner-seat. Then #4 and #6 would be unavailable.
4 available seats are left, and they're #3, #7, #8 and #9.
Step 5: Pick #6 as the 2nd winner-seat. Then #5 and #7 would be unavailable.
4 available seats are left, and they're #3, #4, #8 and #9.
Step 6: Pick #7 as the 2nd winner-seat. Then #6 and #8 would be unavailable.
4 available seats are left, and they're #3, #4, #5 and #9.
Step 7: Pick #8 as the 2nd winner-seat. Then #7 and #9 would be unavailable.
4 available seats are left, and they're #3, #4, #5 and #6.
Step 8: Pick #9 as the 2nd winner-seat. Then #8 would be unavailable.
5 available seats are left, and they're #3, #4, #5, #6 and #7.
In my reasoning, there are
30 ways satisfying the condition which requires
no two adjacent seats will be left empty.
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