MANHATTAN, Kan. – Kansas State was rewarded for a successful season on Sunday night, as the Wildcats earned their eighth bid to the National Invitational Tournament (NIT) and first since 2009.
K-State (19-14, 8-10 Big 12) received an at-large bid and will play No. 3 seed Iowa (18-14, 10-10 Big Ten) in the first round at 8 p.m., CT on Tuesday, March 19 at Carver-Hawkeye Arena in Iowa City, Iowa. The winner will advance to play the winner of No. 2 seed and future Big 12 member Utah (19-14, 9-11 Pac-12) and Big West regular-season champion UC Irvine (24-9, 17-3 Big West) on Saturday, March 23 or Sunday, March 24.
The Wildcats are joined in the 8-team bracket by top-seed Villanova (18-15), 2-seed Utah (19-14), 3-seed Iowa (18-14) and 4-seed UCF (17-15), which will all host first-round games. In addition to the games between K-State and Iowa and Utah and UC Irvine, Villanova will host VCU (22-13) and UCF will host South Florida (24-7).
K-State is making its 40th postseason appearance, which includes 32 in the NCAA Tournament and eight in the Postseason NIT. The Wildcats have now advanced to the postseason 13 times in the last 18 seasons (10 trips to the NCAA Tournament and three to the NIT). It is the second straight postseason bid under head coach Jerome Tang.
The Wildcats were one of 20 at-large selections to the NIT, as conference mates Cincinnati (20-14) and UCF (17-15) received the Big 12's two automatic bids as the teams with the highest NET ratings.