The 8 people at a party shook hands exactly once with each of the others before the ninth person arrived. The ninth person then shook hands with some of these 8 people. A total of 32 handshakes took place. With how many people did the ninth person shake hands?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Label the first 8 people at the party as A, B, C, D, E, F, G, and H.

Then A shakes hands with the 7 others, B accounts for 6 more handshakes (since we have already included B shaking hands with A), C accounts for 5 more handshakes (since we have already included $\mathrm{C}$ shaking hands with A and B), D accounts for 4 more handshakes, E accounts for 3 more handshakes, $\mathrm{F}$ accounts for 2 more handshakes, and $\mathrm{G}$ accounts for 1 more handshake. (H's handshakes with each of the others have already been included.)

So there are $7+6+5+4+3+2+1=28$ handshakes which take place before the ninth person arrives.

Since there are a total of 32 handshakes which take place, then the ninth person shakes hands with $32-28=4$ people.

ANswer: (B)